Math Jams

copeland 2014-02-05 18:47:31
The Math Jam will start at 7:00 PM Eastern / 4:00 PM Pacific.

copeland 2014-02-05 18:47:35
The classroom is moderated, meaning that you can type into the classroom, but these comments will not go directly into the room.

copeland 2014-02-05 18:53:39
Please do not ask about administrative aspects of the contests, and please do not ask me to speculate about the results.

ws0414 2014-02-05 18:54:00
how about passing scores?

copeland 2014-02-05 18:54:00
Please do not ask about administrative aspects of the contests, and please do not ask me to speculate about the results.

arpanliku 2014-02-05 18:54:38
Hello, Mr. Copeland.

copeland 2014-02-05 18:54:40
Hi there!

copeland 2014-02-05 18:56:37
Welcome to the 2014 AMC 10A/12A Math Jam!

copeland 2014-02-05 18:56:43
We will be starting a minute or 2 after 7pm ET.

copeland 2014-02-05 18:57:03
I will explain the structure of the Math Jam shortly.

kajay88 2014-02-05 18:57:36
so we cannot ask you about the cutoff score?

copeland 2014-02-05 18:57:39
You can, but it won't help you. I won't answer.

jigglypuff 2014-02-05 18:58:25
do you know what the cutoff score is?

copeland 2014-02-05 18:58:26
Please do not ask about administrative aspects of the contests, and please do not ask me to speculate about the results.

AkshajK 2014-02-05 18:58:36
what is your stance on rainbows?

copeland 2014-02-05 18:58:36
Rainbows rule.

VinTheSad 2014-02-05 18:59:05
peanut butter?

copeland 2014-02-05 18:59:14
Yep. Peanut butter and rainbows are definitely my thing.

copeland 2014-02-05 18:59:16
How'd you know?

mathman121 2014-02-05 19:00:09
What's your favorite topic in math? (algebra, geometry, number theory, etc.)

adno 2014-02-05 19:00:09
i hate peanut butter

copeland 2014-02-05 19:00:11
Lots of very good, cheap calories. How could you hate it?

pi37 2014-02-05 19:01:02
Geometry has lots of good cheap calories

copeland 2014-02-05 19:01:05
That's right.

bpnyc 2014-02-05 19:01:49
are we starting now?

Art123 2014-02-05 19:01:49
are we starting?

copeland 2014-02-05 19:01:51
Don't mind if I do!

copeland 2014-02-05 19:01:56
Welcome to the 2014 AMC 10A/12A Math Jam!

copeland 2014-02-05 19:01:56
I'm Jeremy Copeland, and I'll be leading our discussion tonight.

copeland 2014-02-05 19:01:59
I'm the school director here at AoPS. That means when something goes wrong, I either get yelled at or have to yell at someone else. Before AoPS, I was an instructor at MIT, and before that I got my Ph.D. from the University of Chicago. Before that I was an undergrad at Reed College and going back even further, I can't really remember. I used to have hobbies, but I'm a parent now, so those days are all over.

copeland 2014-02-05 19:02:09
Before we get started I would like to take a moment to explain our virtual classroom procedures to those who have not previously participated in a Math Jam or one of our online classes.

copeland 2014-02-05 19:02:17
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.

copeland 2014-02-05 19:02:22
This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.

copeland 2014-02-05 19:02:36
There are a lot of students here! As I said, only (a fraction of the) well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!

copeland 2014-02-05 19:02:40
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the prerequisite material for every problem as we go. Another difference between tonight and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask. We usually do in our classes, but we have a large number of students tonight! So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!

AlanMTuring 2014-02-05 19:02:59
How can you type that fast

copeland 2014-02-05 19:03:00
I have 2 keyboards.

copeland 2014-02-05 19:03:09
We have 2 scheduled assistants tonight: Jessica Hyde (i, imaginary) and Benjamin Engwall bluecarneal.

copeland 2014-02-05 19:03:10
Jessica started her career in math education in middle school, helping her classmates with their homework. From there she moved on to tutoring with the Math Honors Society in high school and taking courses at the University of Connecticut. In 2014 she will enroll at the Massachusetts Institute of Technology as a physics major, although she adores physics and mathematics equally and wants it to be known that she does not play favorites between the two. In her spare time she attempts to knit and (fruitlessly) to teach her cat and two rats to obey her authority.

copeland 2014-02-05 19:03:11
Benjamin is a Junior in college who enjoys studying both mathematics and computer science. He has been an active member of the AoPS community for over 5 years, and has enjoyed the opportunities he has had as Moderator, Grader, and TA. In his spare time, he enjoys swimming, golfing, and creating nifty Python programs.

bluecarneal 2014-02-05 19:03:39
Hi!

copeland 2014-02-05 19:03:43
We also have a guest assistant who happened to drop in: Alyssa Zisk (baozhale). Say hi to her, too!

i, imaginary 2014-02-05 19:03:44
Hello!

baozhale 2014-02-05 19:03:55
Hi

simon1221 2014-02-05 19:04:13
hi

aburules 2014-02-05 19:04:13
I am form

adno 2014-02-05 19:04:13
hi

simranK 2014-02-05 19:04:13
hello!

1915933 2014-02-05 19:04:13
Hello!

sunny2000 2014-02-05 19:04:13
hi

ZZmath9 2014-02-05 19:04:13
Hi!

AlanMTuring 2014-02-05 19:04:13
Hi!

APSFUNJJJ 2014-02-05 19:04:13
hi

mualphatheta 2014-02-05 19:04:13
Hi!

1023ong 2014-02-05 19:04:13
nice to meet you all

LukeCell 2014-02-05 19:04:13
Hello

MrMerchant 2014-02-05 19:04:13
Hi!

NikhilP 2014-02-05 19:04:13
Hi

15Pandabears 2014-02-05 19:04:13
hi!

pranavkrishna 2014-02-05 19:04:13
Hello

jigglypuff 2014-02-05 19:04:13
hi

PiesAreSquared 2014-02-05 19:04:13
Hi!

AbsoluteFriend 2014-02-05 19:04:13
Hi!

IsabeltheCat 2014-02-05 19:04:13
Hellow!

ninjataco 2014-02-05 19:04:13
Hello!

ljiang528 2014-02-05 19:04:13
hi

DaChickenInc 2014-02-05 19:04:13
Hello

Ericaops 2014-02-05 19:04:13
Hello

tigergirl 2014-02-05 19:04:13
Hello

lucylai 2014-02-05 19:04:13
hi

poweroftwo 2014-02-05 19:04:13
Hello!

nahor123 2014-02-05 19:04:13
Hi

itsqueencaro 2014-02-05 19:04:13
hi benjamin and jessica!

kbird 2014-02-05 19:04:13
$Hi!$

mathnerd101 2014-02-05 19:04:13
Hello!

copeland 2014-02-05 19:04:20
They can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the incredibly large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over is more likely to annoy us than to get it answered faster, so please, just ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.

copeland 2014-02-05 19:04:25
Please also remember that the purpose of this Math Jam is to work through the solutions to AMC problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. So please, when a question is posted, do not simply respond with the final answer. That's not why we're here. We're going to work through the problems step-by-step, and comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, won't be posted.

copeland 2014-02-05 19:04:35
We will work the last 5 problems from the AMC 10A, then the last 5 problems from the AMC 12A. Two of these problems are the same, 10A Problem 25 and 12A Problem 22. We'll only solve that problem once.

copeland 2014-02-05 19:04:48
We usually promise to chat about other problems on the test, but honestly, I tend to meander and most of us will be exhausted by the end of these 9 problems.

copeland 2014-02-05 19:05:00
Oh, and there's a metric ton of people here tonight (375 and rising). I really don't like saying it, but we're probably going to miss some of the things that some of you say. (Especially during the geometry problems - sheesh.) Please forgive me in advance. That doesn't happen in our classes.

distortedwalrus 2014-02-05 19:05:20
I'm already exhausted

copeland 2014-02-05 19:05:21
You should see it from my side. . .

copeland 2014-02-05 19:05:30
Let's get started, huh?

2kev111 2014-02-05 19:05:50
ok

Anno 2014-02-05 19:05:50
Okay

Turtle 2014-02-05 19:05:50
sure.

rmflute 2014-02-05 19:05:50
yes

ferrastacie 2014-02-05 19:05:50
yes!

mualphatheta 2014-02-05 19:05:50
Yay!

6stars 2014-02-05 19:05:50
Aye

kj2002 2014-02-05 19:05:50
Yeah

dli00105 2014-02-05 19:05:50
yes!!

mathman121 2014-02-05 19:05:50
YEAH!

Pythonprogrammer 2014-02-05 19:05:50
yes!

2kev111 2014-02-05 19:05:50
i cant wait to begin!

copeland 2014-02-05 19:05:56
21. Positive integers $a$ and $b$ are such that the graphs of $y=ax + 5$ and $y=3x + b$ intersect the $x$-axis at the same point. What is the sum of all possible $x$-coordinates of these points of intersection?
$\phantom{peekaboo!}$
$\text{(A) } {-20} \quad \text{(B) } {-18} \quad \text{(C) } {-15} \quad \text{(D) } {-12} \quad \text{(E) } {-8}$

copeland 2014-02-05 19:06:03
What does it mean for $y=ax+5$ and $y=3x+b$ to have the same $x$-intercept?

njaladan 2014-02-05 19:07:18
They both have the same x-value when y=0.

jungyeon 2014-02-05 19:07:18
they intersect at one point with y=0

mathwizard888 2014-02-05 19:07:18
when y is 0, the x-coordinates are equal

copeland 2014-02-05 19:07:28
That's good. Can we make that into equations?

SuperSnivy 2014-02-05 19:07:57
ax+5=3x+b=0

chenmichael9 2014-02-05 19:07:57
$ax+5=3x+b$

mathwizard888 2014-02-05 19:07:57
ax+5=0 and 3x+b=0 have the same solution

Jessicatian 2014-02-05 19:07:57
ax+5=3x+b=0

alex31415 2014-02-05 19:07:57
ax+5=0 and 3x+b=0 for positive integers a,b

copeland 2014-02-05 19:08:00
Right.

copeland 2014-02-05 19:08:01
This means that there's a specific $x$ such that for our $a$ and $b,$
\begin{align*}
0&=ax+5\\
0&=3x+b.
\end{align*}

copeland 2014-02-05 19:08:03
What should we do now?

MathematicsOfPi 2014-02-05 19:08:54
Solve for $x$ in both equations

thkim1011 2014-02-05 19:08:54
solve for x and let them equal

coldsummer 2014-02-05 19:08:54
solve for a and b

Smokkala 2014-02-05 19:08:54
solve for a and b

0_o 2014-02-05 19:08:54
add/subtract equations from each other

mathmaster2012 2014-02-05 19:08:54
solve for a in terms of b

copeland 2014-02-05 19:09:03
That's good, let's try solving these equations.

copeland 2014-02-05 19:09:21
We have 2 equations in 3 unknowns, so we should probably try to eliminate something. What variable don't we want?

ScottBusche 2014-02-05 19:09:37
x

ChenthuranA 2014-02-05 19:09:37
x

surajsreddy 2014-02-05 19:09:37
x

sharonmath 2014-02-05 19:09:37
x

Sesquipedalian 2014-02-05 19:09:37
x

jdew192837 2014-02-05 19:09:37
x

BobCat128 2014-02-05 19:09:37
x

copeland 2014-02-05 19:09:42
Let's eliminate $x$ from the equations. That will give us something Diophantine (meaning an integer equation) in terms of $a$ and $b.$

copeland 2014-02-05 19:09:44
There are lots of ways to do this, and probably several of you want to substitute, but if we use elimination we won't have any nasty denominators. What should we multiply the equations by?

ninjataco 2014-02-05 19:10:17
the first one by 3, second one by a

swe1 2014-02-05 19:10:17
top one by 3 and bottom by a

mjoshi 2014-02-05 19:10:17
3, and a

Tuxianeer 2014-02-05 19:10:17
3 and a respectively

simpleasthat 2014-02-05 19:10:17
3 and -a

flyrain 2014-02-05 19:10:17
3 and a

mjoshi 2014-02-05 19:10:17
the first one by 3, the second one by a?

copeland 2014-02-05 19:10:20
If we multiply the first equation by $3$ and the second equation by $a$ then we can subtract to eliminate $x.$

copeland 2014-02-05 19:10:25
\begin{align*}
0&=3ax+15\\
0&=3ax+ab\\
\hline
0&=15-ab.
\end{align*}

copeland 2014-02-05 19:10:28
Therefore $ab=15.$

copeland 2014-02-05 19:10:30
What are the possible values of $a?$

willwang123 2014-02-05 19:10:50
1 3 5 15

eyux 2014-02-05 19:10:50
1, 3, 5, 15

etothei 2014-02-05 19:10:50
1,3,5,15

fz0718 2014-02-05 19:10:50
1,3,5,15

VinTheSad 2014-02-05 19:10:50
a can equal, 1, 3, 5, or 15

fadebekun 2014-02-05 19:10:50
All factors of $15$ which are $1,3,5,15$

blueferret 2014-02-05 19:10:50
1,3,5,15

ajoy 2014-02-05 19:10:50
1,3,5,15

golden_ratio 2014-02-05 19:10:50
1,3,15

Emmettshell 2014-02-05 19:10:50
1,3,5,15

pranavkrishna 2014-02-05 19:10:50
1,3,5,15

copeland 2014-02-05 19:10:54
We can have $a=1,$ $3,$ $5,$ or $15.$

copeland 2014-02-05 19:10:57
But now we need $x.$ When we solve the first equation for $x$ we get \[x=\frac{-5}a.\]

copeland 2014-02-05 19:10:58
What is the sum of all possible values of $x?$

jeremylu 2014-02-05 19:11:20
-8

Vishnu09 2014-02-05 19:11:20
-8

UrInvalid 2014-02-05 19:11:20
-8

ws5188 2014-02-05 19:11:20
-8

tangaroo 2014-02-05 19:11:20
-8

jigglypuff 2014-02-05 19:11:20
-8

maxplanck 2014-02-05 19:11:20
-8

Offendo 2014-02-05 19:11:20
-8

1915933 2014-02-05 19:11:20
(E)-8

copeland 2014-02-05 19:11:24
We add\[\frac{-5}1+\frac{-5}3+\frac{-5}5+\frac{-5}{15}=-5-\frac53-1-\frac13=\boxed{-8}.\]

copeland 2014-02-05 19:11:25
The answer is (E).

copeland 2014-02-05 19:11:33
Good. I hope everyone's warmed up.

mualphatheta 2014-02-05 19:11:45
yay!

one down

copeland 2014-02-05 19:12:02
Yep. We're working through the AMC10 right now in order from 21 through 25.

copeland 2014-02-05 19:12:04
22. In rectangle $ABCD,$ $AB = 20$ and $BC =10.$ Let $E$ be a point on $\overline{CD}$ such that $\angle CBE = 15^\circ.$ What is $AE?$
$\phantom{peekaboo!}$
$\text{(A) } \dfrac{20\sqrt{3}}{3} \quad \text{(B) } 10\sqrt{3} \quad \text{(C) } 18 \quad \text{(D) } 11\sqrt{3} \quad \text{(E) } 20$

copeland 2014-02-05 19:12:17

copeland 2014-02-05 19:12:19
Anybody have any ideas here?

trumpeter 2014-02-05 19:12:58
trig?

Tuxianeer 2014-02-05 19:12:58
trig

2kev111 2014-02-05 19:12:58
use trigonometry?

Eunectus 2014-02-05 19:12:58
trig

simon1221 2014-02-05 19:12:58
trig?

jillu1947 2014-02-05 19:12:58
i used trig

simranK 2014-02-05 19:12:58
trigonometry...?

Mathdolphin 2014-02-05 19:12:58
Trig bash

copeland 2014-02-05 19:13:00
Trig bash works, but it's a little slower than the "intended" solution.

jeremylu 2014-02-05 19:13:35
15-75-90 triangle

copeland 2014-02-05 19:13:38
I see it!

copeland 2014-02-05 19:13:41
What can we do wit hthat guy?

connor0728 2014-02-05 19:14:41
split it up

guilt 2014-02-05 19:14:41
split 75 into 15 and 60

lightning23 2014-02-05 19:14:41
Split the 15-75-90 into a 30-60 -90 and an isosceles

NikhilP 2014-02-05 19:14:41
75=15+60

copeland 2014-02-05 19:14:47
The easiest solution I have for this problem relies on a trick you might want to know.

copeland 2014-02-05 19:14:48
The trick is: if you ever see a 15-75-90 triangle, draw a new segment that makes a 30-60-90 triangle inside:

copeland 2014-02-05 19:14:51

copeland 2014-02-05 19:14:57
This is really nice because we have on 30-60-90 triangle and we also have an isosceles triangle. This is just one of those tricks you either know or you don't. It's kind of an unfortunate twist to the problem.

copeland 2014-02-05 19:15:01
Write this in your Big Book of Tricks. It doesn't come up that often, but if a problem has a 15-75-90 triangle, you probably should consider starting here.

copeland 2014-02-05 19:15:05
There actually is a more general idea here. If you're give weird angles, start looking for isosceles triangles. That gives you an opportunity to transport side lengths around and it sometimes allows you to double your angles. You can also solve this problem by creating a nice 75-75-30 isosceles triangle somewhere as well, but that solution is a little too mysterious for me.

copeland 2014-02-05 19:15:10
There actually is a more general idea here. If you're given weird angles, start looking for isosceles triangles. That gives you an opportunity to transport side lengths around and it sometimes allows you to double your angles.

copeland 2014-02-05 19:15:27
Incidentally, we can get another nice triangle that we can use the Angle Bisector Theorem on by placing a new point on $DC$ and forcing $BC$ to be an angle bisector in a 30-60-90 triangle.

copeland 2014-02-05 19:15:30

copeland 2014-02-05 19:15:40
Some of you also suggested this. It leads to a fine solution as well.

copeland 2014-02-05 19:15:53
Back to this triangle:

copeland 2014-02-05 19:15:55

copeland 2014-02-05 19:16:00
Now what?

jdew192837 2014-02-05 19:16:44
Use the ratios 1 2 square root of 3

coldsummer 2014-02-05 19:16:44
let CE = x

guor 2014-02-05 19:16:44
Get side lengths.

blueferret 2014-02-05 19:16:44
call EC x

somepersonoverhere 2014-02-05 19:16:44
calculate EC

UrInvalid 2014-02-05 19:16:44
set ec as x, use special triangle formulae

ChenthuranA 2014-02-05 19:16:44
write everything in terms of EC

copeland 2014-02-05 19:16:50
Let's introduce a variable. I'm going to let $x=EC.$

copeland 2014-02-05 19:16:54

copeland 2014-02-05 19:17:00
What other distances do we get?

Richardq 2014-02-05 19:17:43
EF=2x

CountDown 2014-02-05 19:17:43
EF = 2x

Emmettshell 2014-02-05 19:17:43
EF=2x

CountDown 2014-02-05 19:17:43
CF = x * \sqrt{3}

Vishnu09 2014-02-05 19:17:43
EF: 2X

alex31415 2014-02-05 19:17:43
CF=sqrt3*x, EF=2x, FB=2x

jenmath666 2014-02-05 19:17:43
CF=x sqrt3

Turtle 2014-02-05 19:17:43
CF xsqrt3. EF = 2x

dli00105 2014-02-05 19:17:43
CF=xsqrt(3)

pedronr 2014-02-05 19:17:43
CF is $ x\sqrt{3} $

copeland 2014-02-05 19:17:45
$CF=x\sqrt3.$

copeland 2014-02-05 19:17:45
$EF=FB=2x.$

copeland 2014-02-05 19:17:49

copeland 2014-02-05 19:17:56
Now what?

NumberNinja 2014-02-05 19:18:32
CB=10

carjacker 2014-02-05 19:18:32
x*sqrt(3)+2x=10

swe1 2014-02-05 19:18:32
solve for 2x+xsqrt3=10

DigitalKing257 2014-02-05 19:18:32
$CB = 10 = 2x + x\sqrt{3}$

mjoshi 2014-02-05 19:18:32
x(2+\sqrt{3}) = 10

njaladan 2014-02-05 19:18:32
2x + sqrt(3)*x = 10

droid347 2014-02-05 19:18:32
2x+xsqrt(3)=10

copeland 2014-02-05 19:18:36
Oh, now we can solve for $x$ with $BC.$ We have\[10=BC=BF+FC=2x+x\sqrt3.\] Therefore $x=\dfrac{10}{2+\sqrt3}$.

copeland 2014-02-05 19:18:38
What do we get when we rationalize the denominator?

ZZmath9 2014-02-05 19:19:03
20-10sqrt3

jigglypuff 2014-02-05 19:19:03
20-10sqrt3

mathwizard888 2014-02-05 19:19:03

zoroark12345 2014-02-05 19:19:03
20-10/sqrt{3}

simranK 2014-02-05 19:19:03
20-10sqrt{3}

scgorantla 2014-02-05 19:19:03
20-10sqrt3

suli 2014-02-05 19:19:03
20 - 10sqrt(3)

mualphatheta 2014-02-05 19:19:03
20-10sqrt(3)

trumpeter 2014-02-05 19:19:03
20-10sqrt(3)

ninjataco 2014-02-05 19:19:03
$20 - 10\sqrt{3}$

jeremylu 2014-02-05 19:19:03
$20-10\sqrt{3}$

copeland 2014-02-05 19:19:06
\[x=\frac{10}{2+\sqrt3}\cdot\frac{2-\sqrt3}{2-\sqrt3}=\frac{10(2-\sqrt3)}{4-3}=20-10\sqrt3.\]

copeland 2014-02-05 19:19:09
Now what should we find?

mathmaster2012 2014-02-05 19:19:48
DE and pythgagorean

chying1 2014-02-05 19:19:48
DE

zoroark12345 2014-02-05 19:19:48
DE for pythagorean theorem

mathmaster2012 2014-02-05 19:19:48
DE and pythagorean

DrMath 2014-02-05 19:19:48
DE and pythagoras for AE

want2learn 2014-02-05 19:19:48
ED

buzzyun 2014-02-05 19:19:48
DE = 20 -x

copeland 2014-02-05 19:19:55
Since we have $AD=10,$ all we need is $DE$ and the Pythagorean Theorem to get $AE.$ What is $DE?$

Mathlete25 2014-02-05 19:20:19
DE=10\sqrt{3}

jiyer99 2014-02-05 19:20:19
10sqrt3

JixuanW 2014-02-05 19:20:19
DE = 10rt(3)

15Pandabears 2014-02-05 19:20:19
10sqrt3

surajsreddy 2014-02-05 19:20:19
10sqrt(3)

AndrewK 2014-02-05 19:20:19
10 root 3

ZekromReshiram 2014-02-05 19:20:19
10sqrt3

ajc01 2014-02-05 19:20:19
10sqrt3

mathwizard888 2014-02-05 19:20:19

copeland 2014-02-05 19:20:22
This is nice, \[DE=DC-EC=20-(20-10\sqrt3)=10\sqrt3.\]

copeland 2014-02-05 19:20:23
Here's our diagram:

copeland 2014-02-05 19:21:08

pedronr 2014-02-05 19:21:50
ADE is a 30-60-90 triangle

25120012jsy21 2014-02-05 19:21:50
And now we use the pythagorean theorem to get (E), 20?

kajay88 2014-02-05 19:21:50
AE now=20

fadebekun 2014-02-05 19:21:50
So we have $AE=\sqrt{(10\sqrt{3})^2+100}=\sqrt{400}=\boxed{20}$

MathematicsOfPi 2014-02-05 19:21:50
so $\triangle {ADE}$ is also 30-60-90 and we can use the ratios or Pythagorean Theorem to find AE

blueferret 2014-02-05 19:21:50
AE=20

IsabeltheCat 2014-02-05 19:21:50
AE=20 by 30-60-90

WolfOfAtlantis 2014-02-05 19:21:50
30 60 90

pl210741 2014-02-05 19:21:50
30-60-90 triangle

Eunectus 2014-02-05 19:21:50
hey is that a 30-60-90

zacchro 2014-02-05 19:21:50
AE=;$\boxed{20}$

sunny2000 2014-02-05 19:21:50
30, 60, 90 triangle

yuunderstand168 2014-02-05 19:21:50
and that's a 30-60-90 triangle (DAE)

Tuxianeer 2014-02-05 19:21:50
$(\text{E})\ 20$

copeland 2014-02-05 19:21:55
Look! It's a 30-60-90 triangle. We don't even need the Pythagorean Theorem. The hypotenuse is $2\cdot10=\boxed{20}.$ The answer is (E).

copeland 2014-02-05 19:21:58
Incidentally, if you drew a very big, very clear diagram here, you might be able to see either that $\triangle ADE$ is 30-60-90 or that $\triangle AEB$ is isosceles. Either one gives you a good opportunity to guess the answer if you don't see the trick.

copeland 2014-02-05 19:22:12
Phew. 2 down.

laegolas 2014-02-05 19:22:25
I didn't know any of the theorems. I used a ruler and got the answer correct!

fz0718 2014-02-05 19:22:25
If you drew a clear diagram, you could've just used a ruler and got it...

Smokkala 2014-02-05 19:22:25
Somebody used their ruler and found the scale and got 20

copeland 2014-02-05 19:22:29
Absolutely.

mualphatheta 2014-02-05 19:22:45
Finger measurements = Reliable.

copeland 2014-02-05 19:22:46
Depends on your fingers.

copeland 2014-02-05 19:22:48
23. A rectangular piece of paper whose length is $\sqrt{3}$ times the width has area $A.$ The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area $B.$ What is the ratio $B:A?$
$\phantom{peekaboo!}$
$\text{(A) } 1:2 \quad \text{(B) } 3:5 \quad \text{(C) } 2:3 \quad \text{(D) } 3:4 \quad \text{(E) } 4:5$

copeland 2014-02-05 19:23:00

copeland 2014-02-05 19:23:03
Just so we're in the same place on this one, we are going to fold the bottom right corner up and to the left.

copeland 2014-02-05 19:23:24
It probably helps to have some actual scale here. What should we pick for the length of the short edge?

va2010 2014-02-05 19:23:54
\sqrt{3}

ScottBusche 2014-02-05 19:23:54
sqrt3?

LightningX48 2014-02-05 19:23:54
sqrt(3)

DaChickenInc 2014-02-05 19:23:54
$\sqrt{3}$

guilt 2014-02-05 19:23:54
sqrt3

abishek99 2014-02-05 19:23:54
sqrt3

GeorgCantor 2014-02-05 19:23:54
root 3?

copeland 2014-02-05 19:23:57
Since the long edge is trisected, we should make that 3 and the short edge will be $\sqrt3.$

copeland 2014-02-05 19:24:02
It doesn't matter, but I like this choice.

copeland 2014-02-05 19:24:08

copeland 2014-02-05 19:24:13
Now what do you notice?

UrInvalid 2014-02-05 19:24:37
30-60-90

hexagram 2014-02-05 19:24:37
30-60-90!

2kev111 2014-02-05 19:24:37
it is a 30 60 90 triangle

swe1 2014-02-05 19:24:37
30-60-90 triangle

Nitzuga 2014-02-05 19:24:37
30-60-90

LMA 2014-02-05 19:24:37
30-60-90 triangle!

ChenthuranA 2014-02-05 19:24:37
30-60-90

Eudokia 2014-02-05 19:24:37
30-60-90

MasterChief1096 2014-02-05 19:24:37
30-60-90 ratios

Vishnu09 2014-02-05 19:24:37
30-60-90 triangles

copeland 2014-02-05 19:24:40
Every interesting angle in this diagram is in the 30-60-90 triangle family.

copeland 2014-02-05 19:24:45

copeland 2014-02-05 19:24:47
We need to figure out where the bottom right corner lands when we fold it. What is the angle it makes with the horizontal axis after the fold?

harvey2014 2014-02-05 19:25:37
60

pickten 2014-02-05 19:25:37
60

ninjataco 2014-02-05 19:25:37
60

dli00105 2014-02-05 19:25:37
60

Anthrax 2014-02-05 19:25:37
60

ingridzhang97 2014-02-05 19:25:37
60

SuperSnivy 2014-02-05 19:25:37
60

rjw98 2014-02-05 19:25:37
120

6stars 2014-02-05 19:25:37
60 degrees?

copeland 2014-02-05 19:25:39
The point lands along a ray from $X$ that is $120^\circ$ counter-clockwise from $B.$ This is 60 degrees from horizontal to the left.

copeland 2014-02-05 19:25:51

copeland 2014-02-05 19:25:58
Since $XB=2,$ it lies 2 units along this ray. What do you notice?

mymathboy 2014-02-05 19:26:48
B to D

checkmate1021 2014-02-05 19:26:48
B => D

xianmingli 2014-02-05 19:26:48
lies on D

pinetree1 2014-02-05 19:26:48
It is at D

etothei 2014-02-05 19:26:48
B IS ON D

hliu70 2014-02-05 19:26:48
B lands on D

DrMath 2014-02-05 19:26:48
it lands on D

2kev111 2014-02-05 19:26:48
b will touch d

copeland 2014-02-05 19:26:53
Since $\angle AXD=60^\circ,$ $AX=1,$ and $XB=2$ we get that $B$ lands on $D$ after the fold. Wow is that going to be convenient!

copeland 2014-02-05 19:26:56

copeland 2014-02-05 19:27:08
Now we just need to compute. What fraction of the rectangle overlaps?

teachm 2014-02-05 19:27:35
1/3

flyingsledge 2014-02-05 19:27:35
1/3

Mathdolphin 2014-02-05 19:27:35
1/3

15Pandabears 2014-02-05 19:27:35
1/3

rmflute 2014-02-05 19:27:35
1/3

flyrain 2014-02-05 19:27:35
1/3

treemath 2014-02-05 19:27:35
1/3

vincenthuang75025 2014-02-05 19:27:35
1/3

njaladan 2014-02-05 19:27:35
1/3

Kevinyang2000 2014-02-05 19:27:35
1/3

mxgo 2014-02-05 19:27:35
1/3

copeland 2014-02-05 19:27:42
The new figure looks something like this with a few suggestive congruent triangles drawn in.

copeland 2014-02-05 19:27:44

copeland 2014-02-05 19:27:46
We can dissect the rectangle into 6 congruent triangles. Two pairs of the triangles will overlap when we fold giving us just four triangles.

copeland 2014-02-05 19:27:52

copeland 2014-02-05 19:28:09
That means 2/6 of the triangles overlap (a larger equilateral triangle).

copeland 2014-02-05 19:28:12
What is the answer?

pedronr 2014-02-05 19:28:37
so the answer is 2/3

joshualee2000 2014-02-05 19:28:37
so the answer is C 2:3

kajay88 2014-02-05 19:28:37
2/3

LightningX48 2014-02-05 19:28:37
so its 2/3 or (C)

brandonbigbrother 2014-02-05 19:28:37
2:3

El_Ectric 2014-02-05 19:28:37
2/3

want2learn 2014-02-05 19:28:37
2:3

kangrui 2014-02-05 19:28:37
2:3, C

jfu 2014-02-05 19:28:37
C

copeland 2014-02-05 19:28:49
The ratio of the new figure's area to the old figure's area is $\dfrac46=\boxed{\dfrac23}.$ The answer is (C).

blueberry7 2014-02-05 19:29:24
number 24...

arpanliku 2014-02-05 19:29:24
Three down! This is fun!

Basking 2014-02-05 19:29:27
3 down!!

copeland 2014-02-05 19:29:31
24. A sequence of natural numbers is constructed by listing the first 4, then skipping one, listing the next 5, skipping 2, listing 6, skipping 3, and, on the $n$th iteration, listing $n+3$ and skipping $n.$ The sequence begins $1,2,3,4,6,7,8,9,10,13$. What is the $500{,}000$th number in the sequence?
$\phantom{peekaboo!}$
$\text{(A) } 996{,}506 \quad \text{(B) } 996{,}507 \quad \text{(C) } 996{,}508 \quad \text{(D) } 996{,}509 \quad \text{(E) } 996{,}510$

copeland 2014-02-05 19:29:36
Oh, goodie. The bookkeeping problem. Let's have at it, then.

copeland 2014-02-05 19:29:43
What should be the first step in any problem like this?

Tuxianeer 2014-02-05 19:30:12
small cases

fcc1234 2014-02-05 19:30:12
Smaller cases?

mjoshi 2014-02-05 19:30:12
patterns?

simon1221 2014-02-05 19:30:12
find a pattern

ninjackbrosky 2014-02-05 19:30:12
grouping

blueferret 2014-02-05 19:30:12
try to find a pattern

suli 2014-02-05 19:30:12
Find patterns

kbird 2014-02-05 19:30:12
pattern-finding

aburules 2014-02-05 19:30:12
find a pattern

sarvottam 2014-02-05 19:30:12
write the first few terms

lightning23 2014-02-05 19:30:12
Find a pattern

BobCat128 2014-02-05 19:30:12
look for a pattern

copeland 2014-02-05 19:30:23
On basically any bookkeeping problem, I promise you that starting off with a nice clear example will help you. It cleans up all the possible off-by-one errors you might have and gives you some nice test cases for all the general things you'll need to say.

copeland 2014-02-05 19:30:24
Here is the first four blocks of our sequence. This is probably more than you would need. The sequence is in black and the skipped numbers are in red.

copeland 2014-02-05 19:30:26
\begin{array}{cccccccccccc}
1,& 2,& 3,& 4,& {\color{red}{5}},& & & & &&&\\
6,& 7,& 8,& 9,& 10, & {\color{red}{11}},& {\color{red}{12}},& & &&&\\
13,&14,& 15,& 16,& 17,& 18, & {\color{red}{19}},& {\color{red}{20}},& {\color{red}{21}},&&& \\
22,& 23,& 24,& 25,& 26,& 27,& 28,& {\color{red}{29}},&{\color{red}{30}},&{\color{red}{31}},&{\color{red}{32}},&\ldots\\
\end{array}

copeland 2014-02-05 19:30:33
This example will help us to find a pattern.

copeland 2014-02-05 19:30:35
We want the $500{,}000$th black term in this list. What should we count?

gaussian 2014-02-05 19:31:17
terms in each row

Tuxianeer 2014-02-05 19:31:17
number of numbers skipped

willwang123 2014-02-05 19:31:17
how many terms there are in all

Eudokia 2014-02-05 19:31:17
the reds

ingridzhang97 2014-02-05 19:31:17
the black terms

TheCrafter 2014-02-05 19:31:17
Number of terms skipped

mathwizard888 2014-02-05 19:31:17
black and total

angel27 2014-02-05 19:31:17
The number of black numbers in each row

DrMath 2014-02-05 19:31:17
which iteration it is in

gengkev 2014-02-05 19:31:17
black terms and red terms

mathmaster2012 2014-02-05 19:31:17
4,10,18,28, etc.

copeland 2014-02-05 19:31:23
Let's count the number of black terms in the first $n$ blocks and the total number of terms in the first $n$ blocks. How many black terms are there in the first $n$ blocks?

Nahmid 2014-02-05 19:33:01
n(n+7)/2

UrInvalid 2014-02-05 19:33:01
n(n+7)/2

alex31415 2014-02-05 19:33:01
4+5+6+...+(n+3)

dli00105 2014-02-05 19:33:01
4+5+6+...+n+3

linhhuynh 2014-02-05 19:33:01
n(n+7)/2

copeland 2014-02-05 19:33:04
The first $n$ blocks have \[4+5+6+\cdots+(n+3)\]terms. That is $\dfrac{n(n+7)}2$ total black terms in the first $n$ rows.

copeland 2014-02-05 19:33:11
Now let's find the last (red) number in the $n^{\text{th}}$ block. Before that, how many total numbers are on the $n^{\text{th}}$ row?

jigglypuff 2014-02-05 19:34:10
2n+3

wpk 2014-02-05 19:34:10
2n+3

JFC 2014-02-05 19:34:10
3+2n

MaYang 2014-02-05 19:34:10
2n+3

Turtwig123 2014-02-05 19:34:10
2n+3

harvey2014 2014-02-05 19:34:10
2n+3

copeland 2014-02-05 19:34:13
The first row has 5 terms. The second has 7, the third has 9. The $n^{\text{th}}$ row has $2n+3$ total terms.

copeland 2014-02-05 19:34:15
Now what is the last number in row $n?$

UrInvalid 2014-02-05 19:35:12
n(n+4)

mathwizard888 2014-02-05 19:35:12
n(n+4)

Michael_Huang23 2014-02-05 19:35:12
(n)(n+4)

Nahmid 2014-02-05 19:35:12
n(n+4)

brainiac1 2014-02-05 19:35:12
n^2+4n

copeland 2014-02-05 19:35:15
Since the list starts at 1, the last number in row $n$ is equal to the total number of terms in the first $n$ rows. That is \[5+7+9+\cdots+(2n+3)=\frac{n(2n+8)}2=n(n+4).\]

copeland 2014-02-05 19:35:22
A lot of people are asking where these formulas are coming from.

copeland 2014-02-05 19:35:44
If you have an arithmetic progression that starts with $a$ and ends with $b,$ what is the average value?

AndrewK 2014-02-05 19:36:14
a+b/2

NumberNinja 2014-02-05 19:36:14
(a+b)/2

richard4912 2014-02-05 19:36:14
(a+b)/2

Darn 2014-02-05 19:36:14
(A+B)/2

Mathdolphin 2014-02-05 19:36:14
(a+b)/2

MathematicsOfPi 2014-02-05 19:36:14
$\frac {a+b}{2}$

1023ong 2014-02-05 19:36:14
(a+b)/2

SHARKYBOY 2014-02-05 19:36:14
(a+b)/2

Anthrax 2014-02-05 19:36:14
(a+b)/2

eddy55 2014-02-05 19:36:14
(a+b)/2

willwin4sure 2014-02-05 19:36:14
(a+b)/2

copeland 2014-02-05 19:36:19
The average value will be $\frac{a+b}2.$

copeland 2014-02-05 19:36:30
All you need to do to find the sum is multiply this average by the total number of terms.

copeland 2014-02-05 19:36:55
If we start at 5 and end at 2n+3, the average is $\frac{2n+8}2$ and the number of terms is $n.$

copeland 2014-02-05 19:37:08
This is where we get $\dfrac{n(2n+8)}2.$

copeland 2014-02-05 19:37:18
The last omitted (red) number in row $n$ is $n(n+4).$

copeland 2014-02-05 19:37:33
Now we're looking for the $500{,}000$th number in our sequence, which is the $500{,}000$th black number listed. How do we find out what row that term is in?

Nitzuga 2014-02-05 19:38:32
Solve $n(n+7)/2 = 500000$

UrInvalid 2014-02-05 19:38:32
n(n+7)=1000000, ceil(n)

ingridzhang97 2014-02-05 19:38:32
first set n(n+7)/2 equal to 500,000 to find n

blueferret 2014-02-05 19:38:32
n(n+7)/2>500000

AaaDuhLoo 2014-02-05 19:38:32
n(n+7)/2 = 500,000

checkmate1021 2014-02-05 19:38:32
n(n+7)/2 = 500000

JFC 2014-02-05 19:38:32
Find what value of n, when in n(n+7)/2, exceeds 500000

ingridzhang97 2014-02-05 19:38:32
first set n(n+7)/2 equal to 500,000 to find the iteration/row

copeland 2014-02-05 19:38:37
We want to find the $n$ such that \[\frac{n(n+7)}2<500{,}000\leq\frac{(n+1)(n+8)}2.\]

copeland 2014-02-05 19:38:39
Let's look at $n(n+7)\approx1{,}000{,}000.$ What is an obvious place to start?

Emmettshell 2014-02-05 19:38:59
1000

zoroark12345 2014-02-05 19:38:59
1000

joshualee2000 2014-02-05 19:38:59
1000

ninjackbrosky 2014-02-05 19:38:59
1000

DrMath 2014-02-05 19:38:59
1000

geo31415926 2014-02-05 19:38:59
1000

DuoCapital 2014-02-05 19:38:59
1000

gengkev 2014-02-05 19:38:59
1000!

15Pandabears 2014-02-05 19:38:59
1000

LightningX48 2014-02-05 19:38:59
1000

dli00105 2014-02-05 19:38:59
n=1000

dlin 2014-02-05 19:38:59
1000

matthewzou 2014-02-05 19:38:59
1000

copeland 2014-02-05 19:39:02
If we let $n=1000$ we get $n(n+7)=1000\cdot1007=1{,}007{,}000$ which is about 7000 too much.

copeland 2014-02-05 19:39:08
You could decrease $n$ a few times and find the right row, but here's a quick trick that might shave some seconds off your time:

copeland 2014-02-05 19:39:10
If $n\approx1000$ and we decrease $n$ by 1, about how much do we decrease $n(n+7)?$

Michael_Huang23 2014-02-05 19:39:52
2000

JFC 2014-02-05 19:39:52
2000

dli00105 2014-02-05 19:39:52
2000

Darn 2014-02-05 19:39:52
2000>

NathanV 2014-02-05 19:39:52
2000

willwin4sure 2014-02-05 19:39:52
2000

checkmate1021 2014-02-05 19:39:52
2000?

alex31415 2014-02-05 19:39:52
2000

billgates42 2014-02-05 19:39:52
2000

copeland 2014-02-05 19:39:54
The Binomial Theorem tells us that if $(a-1)(b-1)=ab-a-b+1.$ So if $a$ and $b$ are close to 1000, decreasing both by 1 decreases the product by about $a+b\approx2000.$ Every time we decrease $n$ by 1 we decrease the product by approximately 2000.

copeland 2014-02-05 19:40:03
Therefore, what $n$ should we start with?

poweroftwo 2014-02-05 19:40:28
996

pedronr 2014-02-05 19:40:28
996

ssk9208 2014-02-05 19:40:28
996

Turtwig123 2014-02-05 19:40:28
996

DaChickenInc 2014-02-05 19:40:28
996

CornSaltButter 2014-02-05 19:40:28
996

scgorantla 2014-02-05 19:40:28
996

joshualee2000 2014-02-05 19:40:28
996

BobCat128 2014-02-05 19:40:28
996

copeland 2014-02-05 19:40:32
We know $n=1000$ is about 7000 too big so replacing $n$ by $n-4$ should decrease the product by about $4\cdot2000=8000.$ That is the row before the $5000{,}000$th term.

copeland 2014-02-05 19:40:34
How many terms are in the first 996 rows?

guilt 2014-02-05 19:41:33
996000

UrInvalid 2014-02-05 19:41:33
996000

LightningX48 2014-02-05 19:41:33
996000

Richardq 2014-02-05 19:41:33
996000

mathwizard888 2014-02-05 19:41:33
996000

mxgo 2014-02-05 19:41:33
996000

Nitzuga 2014-02-05 19:41:33
$996,000$

hametm 2014-02-05 19:41:33
n(n+4), which is 996(1000)=996000

copeland 2014-02-05 19:41:35
The first 996 rows have \begin{align*}
\frac{996\cdot(996+7)}2
&=\frac12(1000-4)(1000+3)\\
&=\frac12(1{,}000{,}000-4000+3000-12)\\
&=500{,}000-506
\end{align*}terms.

copeland 2014-02-05 19:41:39
What does that tell us?

IsabeltheCat 2014-02-05 19:42:25
997th row

pedronr 2014-02-05 19:42:25
the 500000th term is in the 997th row

ferrastacie 2014-02-05 19:42:25
its the 997 row

anwang16 2014-02-05 19:42:25
its in the 997th row

copeland 2014-02-05 19:42:32
Yep. Where is it in that row?

mathstar10 2014-02-05 19:43:02
506 of the 997 row

billgates42 2014-02-05 19:43:02
506 more in the next row

Eudokia 2014-02-05 19:43:02
506th term

jigglypuff 2014-02-05 19:43:02
506th number

trumpeter 2014-02-05 19:43:02
the 507th term

mathstar10 2014-02-05 19:43:02
506

mjoshi 2014-02-05 19:43:02
506th

vincenthuang75025 2014-02-05 19:43:02
506th

rmflute 2014-02-05 19:43:02
506th place

angel27 2014-02-05 19:43:02
506th

VinTheSad 2014-02-05 19:43:02
506th place

NikhilP 2014-02-05 19:43:02
506th term

golden_ratio 2014-02-05 19:43:02
506

copeland 2014-02-05 19:43:05
This tells us that we want the 506th term in the 997th row.

copeland 2014-02-05 19:43:06
To finish, we need to know the last (red) number in the 996th row. What is it?

mathwizard888 2014-02-05 19:43:43
996000

bookie331 2014-02-05 19:43:43
996000

Darn 2014-02-05 19:43:43
996000?

mymathboy 2014-02-05 19:43:43
996,000

copeland 2014-02-05 19:43:51
Right. We already computed this.

copeland 2014-02-05 19:43:52
The last red number in row 996 is $996(996+4)=996{,}000.$

copeland 2014-02-05 19:43:57
And what's the final answer?

simon1221 2014-02-05 19:44:23
996506

SuperSnivy 2014-02-05 19:44:23
996506

kbird 2014-02-05 19:44:23
A is the answer!

Tuxianeer 2014-02-05 19:44:23
$\text{(A)} 996,506$

buzzyun 2014-02-05 19:44:23
A

jkyman 2014-02-05 19:44:23
A

kajay88 2014-02-05 19:44:23
a

LogPolarBear 2014-02-05 19:44:23
996506, or $A$

flamesofpi 2014-02-05 19:44:23
A

MathematicsOfPi 2014-02-05 19:44:23
$996,506$

njaladan 2014-02-05 19:44:23
996506

ZekromReshiram 2014-02-05 19:44:23
996506

lightning23 2014-02-05 19:44:23
996506

ZZmath9 2014-02-05 19:44:23
996506

ChenthuranA 2014-02-05 19:44:23
996506

copeland 2014-02-05 19:44:26
The answer is $996{,}000+506=\boxed{996{,}506}.$ (A)

copeland 2014-02-05 19:44:43
Alright, good work.

copeland 2014-02-05 19:44:49
We're moving right along it seems.

copeland 2014-02-05 19:44:59
The next problem is the only overlap between the tests:

copeland 2014-02-05 19:45:15
25. The number $5^{867}$ is between $2^{2013}$ and $2^{2014}.$ How many pairs of integers $(m,n)$ are there such that $1\le m \le 2012$ and
\[5^n < 2^m < 2^{m+2} < 5^{n+1}?\]
$\text{(A) } 278 \quad \text{(B) } 279 \quad \text{(C) } 280 \quad \text{(D) } 281 \quad \text{(E) } 282$

copeland 2014-02-05 19:45:25
This was also Problem 22 on the AMC12.

copeland 2014-02-05 19:45:45
Here's a neat problem. It's not quite clear what the purpose of the first sentence is yet, so let's focus on the second. In words, what are we trying to find?

DrMath 2014-02-05 19:46:41
three powers of 2 in between consecutive powers of 5

jweisblat14 2014-02-05 19:46:41
how many times, when we order all the powers of 2 and 5, we get a 5 2 2 2 5 pattern

Nahmid 2014-02-05 19:46:41
trying to find 3 powers of 2 in between 2 powers of 5

jigglypuff 2014-02-05 19:46:41
3 powers of 2 are between 2 powers of 5

alex31415 2014-02-05 19:46:41
The number of triplets on consecutive powers of 2 that are between the same two powers of 5

wpk 2014-02-05 19:46:41
3 powers of two between consecutive powers of 5

copeland 2014-02-05 19:46:42
We are trying to find the pairs of consecutive powers of 5 that have at least 3 different powers of 2 in between them.

copeland 2014-02-05 19:46:50
Now the target is clear, but I'm not really sure where to start. What should we do?

pl210741 2014-02-05 19:47:34
check small cases?

alex31415 2014-02-05 19:47:34
Try small cases

pedronr 2014-02-05 19:47:34
find small examples

dli00105 2014-02-05 19:47:34
smaller values

willwin4sure 2014-02-05 19:47:34
try some small numbers?

joshualee2000 2014-02-05 19:47:34
find the first number then find a pattern?

anwang16 2014-02-05 19:47:34
Try small cases.

suli 2014-02-05 19:47:34
List some examples

AndrewK 2014-02-05 19:47:34
Find an example case

mjoshi 2014-02-05 19:47:34
test out some values?

JFC 2014-02-05 19:47:34
Find the first solution?

want2learn 2014-02-05 19:47:34
the smallest number m this works for

copeland 2014-02-05 19:47:39
Let's look at the first few. The powers of 5 start \[1,5,25,125,625,3125,\ldots.\]The powers of 2 start\[2,4,8,16,32,64,128,256,512,1024,\ldots.\](We've omitted 1 from our 2's list because of the $m \ge 1$ condition.)

copeland 2014-02-05 19:47:53
That's a little confusing. When I was solving, I found it much easier to put those all in the same list. This is where having two colors of ink really helps.

copeland 2014-02-05 19:47:56
\[{\color{red}{1}}, 2, 4, {\color{red}{5}}, 8, 16, {\color{red}{25}}, 32, 64, {\color{red}{125}}, 128, 256, 512, {\color{red}{625}}, 1024, 2048, {\color{red}{3125}},\ldots.\]

copeland 2014-02-05 19:48:04
What do we notice from our list?

ABCDE 2014-02-05 19:48:51
so 2 or 3 powers of 2 between each power of 5

jweisblat14 2014-02-05 19:48:51
2,2,2,3,2,...

MasterChief1096 2014-02-05 19:48:51
there are 2 or 3 powers of 2 between every two powers of 5

fz0718 2014-02-05 19:48:51
2,2,2,3,2

Sesquipedalian 2014-02-05 19:48:51
2,2,3,2,2,3,2,2,3,2,2,3

hexagram 2014-02-05 19:48:51
only gaps of 2 and 3

blueferret 2014-02-05 19:48:51
theres always 2 powers of 2, but only occasionally 3

chying1 2014-02-05 19:48:51
there is a power of 5 between every 2 or 3 powers of 2

copeland 2014-02-05 19:48:55
For all the powers of 5 (in red) in our list, there's always two powers of 2 between them, except for 125 and 625 that have three power of 2 between them.

copeland 2014-02-05 19:48:57
So we might conjecture that there's always either two or three powers of 2 between every consecutive pairs of powers of 5.

copeland 2014-02-05 19:49:02
Can we prove this?

VinVinB 2014-02-05 19:49:19
Yes.

hexagram 2014-02-05 19:49:19
yes

Tuxianeer 2014-02-05 19:49:19
yes

suli 2014-02-05 19:49:19
Yes

jigglypuff 2014-02-05 19:49:19
yes

mathwizard888 2014-02-05 19:49:19
yes

hshiems 2014-02-05 19:49:19
Yes.

goldentail141 2014-02-05 19:49:19
yes!

Richardq 2014-02-05 19:49:19
yes

copeland 2014-02-05 19:49:23
I love the confidence!

copeland 2014-02-05 19:49:35
Let's just look between two consecutive powers of 5. Suppose the smallest power of 2 in between them is $2^a$ and the largest is $2^b:$
\[ {\color{red}{5^n}} < 2^a < 2^b < {\color{red}{5^{n+1}}} \]

copeland 2014-02-05 19:49:38
What are we interested in?

UrInvalid 2014-02-05 19:50:23
b-a

poweroftwo 2014-02-05 19:50:23
b-a

DrMath 2014-02-05 19:50:23
a-b=1 or 2

mymathboy 2014-02-05 19:50:23
b-a

guilt 2014-02-05 19:50:23
b-a

DaChickenInc 2014-02-05 19:50:23
possible values of b-a

jigglypuff 2014-02-05 19:50:23
b-a

pl210741 2014-02-05 19:50:23
b-a

copeland 2014-02-05 19:50:28
We're interested in how many powers of 2 there can be, so we're interested in $b-a.$

copeland 2014-02-05 19:50:30
How can we get at $b-a$ algebraically?

buzzyun 2014-02-05 19:51:35
dividing

pattycakechichi 2014-02-05 19:51:35
logs

Pythonprogrammer 2014-02-05 19:51:35
divide

simon1221 2014-02-05 19:51:35
log base 2

trumpeter 2014-02-05 19:51:35
logarithms

Nitzuga 2014-02-05 19:51:35
logarithms?

hwl0304 2014-02-05 19:51:35
logs?

Eudokia 2014-02-05 19:51:35
dividing

MathPro1000 2014-02-05 19:51:35
Divide

copeland 2014-02-05 19:51:38
This is a mixed bag here. We'll talk about logarithms in a bit, but since we're solving this in the context of the AMC10, let's skip the idea for a moment.

copeland 2014-02-05 19:51:43
Well, we know that $2^{b-a} = \dfrac{2^b}{2^a}.$

copeland 2014-02-05 19:51:46
And how can we relate that fraction to powers of 5?

copeland 2014-02-05 19:52:43
Let's try to bound this above. I know something about $2^b,$ right?

ninjataco 2014-02-05 19:53:23
less than $5^{n+1}$

suli 2014-02-05 19:53:23
2^b < 5^(n+1)

distortedwalrus 2014-02-05 19:53:23
it's less than 5^{n+1}

VinTheSad 2014-02-05 19:53:23
yeah, 2^b<5^n+1

checkmate1021 2014-02-05 19:53:23
$2^b < 5^(n-1)$

ferrastacie 2014-02-05 19:53:23
2^b < 5^(n+1)

pedronr 2014-02-05 19:53:23
2^b<5^(n+1)

copeland 2014-02-05 19:53:29
Right. We know that $2^b < 5^{n+1},$ so replacing the numerator with the larger $5^{n+1}$ makes the fraction larger.

copeland 2014-02-05 19:53:38
And I know something about $2^a,$ right?

joshualee2000 2014-02-05 19:54:07
bigger than 5^n

lee42 2014-02-05 19:54:07
greater than 5^n

fcc1234 2014-02-05 19:54:07
2a>5^n

golden_ratio 2014-02-05 19:54:07
greater than 5^n

ingridzhang97 2014-02-05 19:54:07
it's larger 5^n

dli00105 2014-02-05 19:54:07
2^a>5^n

flamefoxx99 2014-02-05 19:54:07
It's larger than 5^n

angel27 2014-02-05 19:54:07
It's bigger than 5^n

AndrewK 2014-02-05 19:54:07
Its greater thatn $5^n$

MaYang 2014-02-05 19:54:07
5^n < 2^a

ChenthuranA 2014-02-05 19:54:07
5^n<2^a

copeland 2014-02-05 19:54:10
We also know that $5^n < 2^a,$ so replacing the denominator with the smaller $5^n$ also make the fraction larger.

copeland 2014-02-05 19:54:13
Therefore, \[ 2^{b-a} = \frac{2^b}{2^a} < \frac{5^{n+1}}{5^n} = 5. \]

copeland 2014-02-05 19:54:13
What have we concluded?

ninjataco 2014-02-05 19:55:48
$b-a \leq 2$

DrMath 2014-02-05 19:55:48
2^b-a is at most 5 so b-a<=2

chying1 2014-02-05 19:55:48
b-a<3

ZekromReshiram 2014-02-05 19:55:48
which means the interval can only be 0, 1, or 2

DaChickenInc 2014-02-05 19:55:48
$b-a\le 2$

Kevinyang2000 2014-02-05 19:55:48
b-a<=2

IceFireGold1 2014-02-05 19:55:48
2>=b-a

AopsKevin 2014-02-05 19:55:48
b-a<log2(5)

copeland 2014-02-05 19:55:52
$2^{b-a} < 5$, so $b-a \le 2$.

copeland 2014-02-05 19:55:55
So there can only be at most 3 different powers of 2 between any two consecutive powers of 5.

copeland 2014-02-05 19:56:04
Notice that we haven't ruled out $b=a$ yet.

copeland 2014-02-05 19:56:07
How about the other side of the argument? We had conjectured that there always had to be exactly two or three powers of 2, and we've just proved that there can't be more than three -- how do we show that there can't be fewer than two?

1915933 2014-02-05 19:56:33
Show that b-a>0.

flamefoxx99 2014-02-05 19:56:33
We show that a-b \ge 1

copeland 2014-02-05 19:56:39
Sounds like a noble goal.

copeland 2014-02-05 19:56:51
How do we get an inequality that is reversed?

copeland 2014-02-05 19:57:13
Specifically, we want something in terms of $b$ to be bigger than someting with a 5.

jweisblat14 2014-02-05 19:58:28
2^(b+1)>5

jweisblat14 2014-02-05 19:58:28
and 2^(a-1)<5^n

zoroark12345 2014-02-05 19:58:28
take the largest b and then take 2^(b+1)

pedronr 2014-02-05 19:58:28
2^(b+1)>5^(n+1)

copeland 2014-02-05 19:58:38
Good. Nice.

copeland 2014-02-05 19:58:42
Let's look and the next powers of two immediately before and after those in the middle: that is, we'll extend our inequality a little bit.
\[ 2^{a-1} < {\color{red}{5^n}} < 2^a < 2^b < {\color{red}{5^{n+1}}} < 2^{b+1} \]

copeland 2014-02-05 19:58:52
Now what? Are we in business?

copeland 2014-02-05 19:59:05
What should we divide now?

UrInvalid 2014-02-05 19:59:52
the outer terms

ingridzhang97 2014-02-05 19:59:52
2^(b+1) and 2^(a-1)

pedronr 2014-02-05 19:59:52
2^(b+1)/2(a-1)>5

blueferret 2014-02-05 19:59:52
2^(b+1)/2^(a-1)

jiecut 2014-02-05 19:59:52
2^b+1/2^a-1

hametm 2014-02-05 19:59:52
2^(b+1) and 2^(a-1)

VinTheSad 2014-02-05 19:59:52
2^b+1 and 2^a-1

jweisblat14 2014-02-05 19:59:57
divide the powers of 5, and divide the outer powers of 2

tfmtoto 2014-02-05 20:00:00
2^(b+1)/2^(a-1)

copeland 2014-02-05 20:00:02
We can look at the ratio of the new powers of 2, just like we did before.

copeland 2014-02-05 20:00:03
We start with $2^{b-a+2} = \dfrac{2^{b+1}}{2^{a-1}}.$

copeland 2014-02-05 20:00:06
Now what?

lee42 2014-02-05 20:00:54
compare with powers of 5

pedronr 2014-02-05 20:00:54
greater than 5^(n+1)/5^n

hametm 2014-02-05 20:00:54
Bound it from above like before

ingridzhang97 2014-02-05 20:00:54
and that is greater than 5^(n+1)/5^n=5

dli00105 2014-02-05 20:00:54
round top down and bottom up

angel27 2014-02-05 20:00:54
2^b+1/2^a-1 > 5^n+1/5^n

bigc08 2014-02-05 20:00:54
that is greater than 5^n+1/5^n

copeland 2014-02-05 20:00:57
If we replace the numerator with $5^{n+1},$ we make the fraction smaller (since the numerator becomes smaller).

copeland 2014-02-05 20:01:00
And if we replace the denominator with $5^n,$ we also make the fraction smaller (since the denominator becomes larger).

copeland 2014-02-05 20:01:04
Therefore,
\[ 2^{b-a+2} = \frac{2^{b+1}}{2^{a-1}} > \frac{5^{n+1}}{5^n} = 5. \]

copeland 2014-02-05 20:01:07
What have we concluded?

DrMath 2014-02-05 20:01:52
b-a>0

Tuxianeer 2014-02-05 20:01:52
b-a>0

ChenthuranA 2014-02-05 20:01:52
b-a+2>2

joshualee2000 2014-02-05 20:01:52
b-a+2>2

MaYang 2014-02-05 20:01:52
b-a+2>=2

mathbeida 2014-02-05 20:01:52
b-a+2 > 2

goldentail141 2014-02-05 20:01:52
a cannot be equal to b

hamup1 2014-02-05 20:01:52
b-a+2>2, or b-a>0

djmathman 2014-02-05 20:01:52
That $b-a+2\geq 3\implies b-a\geq 1$

copeland 2014-02-05 20:01:59
and also. . .

MATHCOUNTSmath 2014-02-05 20:02:03
that math is awesome

copeland 2014-02-05 20:02:05
$2^{b-a+2} > 5$, so $b-a+2 \ge 3$, and thus $b-a \ge 1$.

copeland 2014-02-05 20:02:18
We've shown that if $2^a$ and $2^b$ are the smallest and largest powers of 2 between any two consecutive powers of 5, then $1 \le b-a \le 2$.

copeland 2014-02-05 20:02:27
What does that mean?

ferrastacie 2014-02-05 20:03:13
b-a = 1 or 2

JFC 2014-02-05 20:03:13
There are always 2 or 3 powers of 2 between powers of 5

simon1221 2014-02-05 20:03:13
there are 2 or 3 powers of 2

hamup1 2014-02-05 20:03:13
There is either 2 or 3 powers of two in between two consecutive powers of 5

VinTheSad 2014-02-05 20:03:13
there must be either 2 or 3 powers of 2 between each power of 5

pedronr 2014-02-05 20:03:13
there are either 2 or 3 consecutive powers of 2 between each two powers of 5

LMA 2014-02-05 20:03:13
there can only be 2 or 3 integer powers of 2 between any 2 consecutive powers of 5

AlcumusGuy 2014-02-05 20:03:13
we've proven the conjecture that there must be either 2 or 3 powers of 2 between two consecutive powers of 5

UltimateChampion1235 2014-02-05 20:03:13
there can be 2-3 powers of 2 between cons. powers of 5

wpk 2014-02-05 20:03:13
2 or 3 powers of 2 between consecutive powers of 5

mxgo 2014-02-05 20:03:13
there can only be 2-3 powers of 2

checkmate1021 2014-02-05 20:03:13
we can only have 2 or 3 powers of 2 petween two consecutive powers of five

1915933 2014-02-05 20:03:13
Only two or three consecutive powers of 2 can be between two consecutive powers of 5.

abishek99 2014-02-05 20:03:13
there are 2 or 3 powers of 2 between each pair of consecutive powers of 5

firemike 2014-02-05 20:03:13
there can only two or three terms between the two powers of 5

copeland 2014-02-05 20:03:27
Putting these two results together tell us that every consecutive pair of powers of 5 has exactly two or three powers of 2 between them. We've proved our conjecture.

copeland 2014-02-05 20:03:30
Great! How do we use this information to solve the problem?

copeland 2014-02-05 20:04:32
A lot of you are suggesting that we find a pattern now.

copeland 2014-02-05 20:04:42
You can actually PROVE that there is no pattern.

coldsummer 2014-02-05 20:04:57
we want the number of cases in which there are three powers of two in between two consecutive powers of 5

fadebekun 2014-02-05 20:04:57
We want the number of intervals with 3 powers of 2.

copeland 2014-02-05 20:05:01
All we want is this.

copeland 2014-02-05 20:05:06
Remember the barnyard problem?

copeland 2014-02-05 20:05:44
If I have a farm and there are 100 animals and they have 150 feet, how many animals are sheep and how many are cows?

zacchro 2014-02-05 20:05:54
I can guess--x cows with 4 legs and y chickens with 2?

copeland 2014-02-05 20:06:12
So now we have a happy analogy.

copeland 2014-02-05 20:06:19
Oh, wait.

copeland 2014-02-05 20:06:22
That doesn't work.

willwin 2014-02-05 20:06:31
sheep and cows both have 4 legs though

copeland 2014-02-05 20:06:36
Sheep and cows happen to have the same number of legs.

copeland 2014-02-05 20:06:55
Let's try something else. How about alpacas and flamingos?

copeland 2014-02-05 20:07:06
What are our flamingos in this example?

angel27 2014-02-05 20:07:36
The 2s

tekgeek 2014-02-05 20:07:36
the number of powers of 2

want2learn 2014-02-05 20:07:36
an interval with 2

anonymous0 2014-02-05 20:07:36
2 powers of 2

Offendo 2014-02-05 20:07:36
2 powers

Lord.of.AMC 2014-02-05 20:07:36
the cases in which there are 2 powers of 2 in between

copeland 2014-02-05 20:07:43
Yes, and the alpacas are the cases of 3.

copeland 2014-02-05 20:07:57
We just need to count the consecutive powers of 5 that have three powers of 2 between them. (Each such pair will contribute exactly one ordered pair $(m,n)$ to our final count.)

copeland 2014-02-05 20:08:05
Hmmm...what data haven't we used yet?

blueferret 2014-02-05 20:09:01
the first sentence

suli 2014-02-05 20:09:01
First sentence

AaaDuhLoo 2014-02-05 20:09:01
The first sentence!

lee42 2014-02-05 20:09:01
the first sentence

DVA6102 2014-02-05 20:09:01
the first sentence

UrInvalid 2014-02-05 20:09:01
the first sentence

Nitzuga 2014-02-05 20:09:01
The first sentence!

jkyman 2014-02-05 20:09:01
the first sentence

copeland 2014-02-05 20:09:08
We haven't used $2^{2013} < 5^{867} < 2^{2014}$ yet. How does that figure in?

goodbear 2014-02-05 20:09:50
5^867 is between 2^2013 and 2^2014

bigc08 2014-02-05 20:09:50
only 866 intervals

ingridzhang97 2014-02-05 20:09:50
the highest power of 5 that we can consider is 867, and there are a total of 2013 powers of 2

Richardq 2014-02-05 20:09:50
it shows that n cannot be 867

jweisblat14 2014-02-05 20:09:50
there are 2013 legs and 867 animals - how many have 3 feet?

zoroark12345 2014-02-05 20:09:50
tells us that there are 2013 powers of 2 under 5^867

copeland 2014-02-05 20:09:53
It tells us that $5^{867}$ is the largest power of 5 that we need to look at (since $m \le 2012$).

copeland 2014-02-05 20:09:55
So we're looking at the powers of 2 that lie in:
\[ 5^0 < \cdots < 5^1 < \cdots < 5^2 < \cdots < \cdots\cdots < \cdots < 5^{866} < \cdots < 5^{867} \]

copeland 2014-02-05 20:09:59
That gives us 867 possible pairs of consecutive powers of 5.

copeland 2014-02-05 20:10:03
Let's finish our sequence. Now we know that it ends

copeland 2014-02-05 20:10:04
\[\ldots,2^{2012},2^{2013},{\color{red}{5^{867}}},2^{2014}\]

copeland 2014-02-05 20:10:16
And how many powers of 2 do we care about?

fadebekun 2014-02-05 20:10:51
2013 powers

ChenthuranA 2014-02-05 20:10:51
2013

Darn 2014-02-05 20:10:51
2013

alex31415 2014-02-05 20:10:51
2013

Showpar 2014-02-05 20:10:51
2013

Eudokia 2014-02-05 20:10:51
2013

pedronr 2014-02-05 20:10:51
2013 powers

coldsummer 2014-02-05 20:10:51
2013

Emmettshell 2014-02-05 20:10:51
2013

hametm 2014-02-05 20:10:51
2013

brainiac1 2014-02-05 20:10:51
2013

mathsd 2014-02-05 20:10:51
2013

copeland 2014-02-05 20:10:55
We start at $2^1$ and end at $2^{2013},$ so there are 2013 of them.

copeland 2014-02-05 20:11:08
That's where having the example protects us from off-by-one errors.

copeland 2014-02-05 20:11:11
And we know that there are either two of three of them in each of the 867 gaps between powers of 5.

copeland 2014-02-05 20:11:28
If we go ahead and allot 2 legs to each animal,, how many legs are left over?

Tuxianeer 2014-02-05 20:12:12
279

joshualee2000 2014-02-05 20:12:12
279

Ericaops 2014-02-05 20:12:12
279

jigglypuff 2014-02-05 20:12:12
279 (B)

coldsummer 2014-02-05 20:12:12
279

UrInvalid 2014-02-05 20:12:12
279

DrMath 2014-02-05 20:12:12
279

dli00105 2014-02-05 20:12:12
279

flamesofpi 2014-02-05 20:12:12
279

kgator 2014-02-05 20:12:12
279

ninjataco 2014-02-05 20:12:12
279

copeland 2014-02-05 20:12:15
There are $2013 - 2(867) = 2013 - 1734 = 279$ remaining. That's how many of the gaps that must have a third power of 2 in order to fit all 2013 of them in there.

GeorgCantor 2014-02-05 20:12:23
2*the number of alpacas

ksun48 2014-02-05 20:12:23
er wait 2 legs to each

copeland 2014-02-05 20:12:39
Yeah, well, no. For the purposes of this exercise, we've modified our alpacas.

lightning23 2014-02-05 20:13:02
279

1915933 2014-02-05 20:13:02
279

mathwizard888 2014-02-05 20:13:02
279

smith_ch 2014-02-05 20:13:02
279

dli00105 2014-02-05 20:13:02
(B)279

copeland 2014-02-05 20:13:05
So there are 279 consecutive powers of 5 with exactly three powers of 2 between them. Our answer is 279, answer (B).

mathnerd101 2014-02-05 20:13:43
Why didn't you use bicycles and tricycles?

copeland 2014-02-05 20:13:45
They are not useful if you want to shear them for sweater yarn.

LMA 2014-02-05 20:14:09
so, are flamingos useful for sweaters???

copeland 2014-02-05 20:14:12
You should see my flamingo sweaters. They're fabulous.

baozhale 2014-02-05 20:14:33
Very feathery

copeland 2014-02-05 20:14:34
This problem was also on the AMC12, and if you solved it there, you might have used a little more machinery.

copeland 2014-02-05 20:14:37
If you're anything like me (and most people aren't), then you instead took the log of everything immediately after seeing the problem. In this point of view, we want to put the powers of two sequence \[\log2,2\log2,3\log2,\ldots,2014\log2\] in between the terms of the sequence \[\log5,2\log5,3\log5,\ldots,867\log5.\]

copeland 2014-02-05 20:14:52
What's cool about those sequences?

pl210741 2014-02-05 20:15:22
arithmetic?

suli 2014-02-05 20:15:22
Arithmetic

MATHCOUNTSmath 2014-02-05 20:15:22
arithmetic

pattycakechichi 2014-02-05 20:15:22
arithmetic

Dragon6point1 2014-02-05 20:15:22
arithmetic?

joshualee2000 2014-02-05 20:15:22
arithmetic

bengals 2014-02-05 20:15:22
arithmetic

simranK 2014-02-05 20:15:22
arithmetic!

MATHCOUNTSmath 2014-02-05 20:15:22
They are arithmetic series

AkshajK 2014-02-05 20:15:22
they are arithmetic sequences

WildTurtle 2014-02-05 20:15:22
Arithmetic sequence

copeland 2014-02-05 20:15:24
These are arithmetic sequences! The 2-sequence is much closer together than the 5-sequence. Anybody see a nice rational approximation to $\dfrac{\log5}{\log2}?$

Tuxianeer 2014-02-05 20:16:19
2013/867

lucylai 2014-02-05 20:16:19
use the first sentence

pi37 2014-02-05 20:16:19
2013.5/867

pattycakechichi 2014-02-05 20:16:19
2014/867?

copeland 2014-02-05 20:16:23
We know for sure that the sequence end around the same place so $865\log5\approx 2014\log2$ and in particular, the ratio of the common differences is \[\frac{\log5}{\log2}\approx\frac{2014}{865}\] which is a little bit more than 2.

copeland 2014-02-05 20:16:29
Now it's quite useful to know that if you have two infinite arithmetic progressions with common ratio between $n$ and $n+1$ then (whenever there are no terms in common) the shorter-period progression appears either $n$ or $n+1$ times in every gap of the longer period one. In our case, we see that we always get 2 or 3 terms in each gap.

copeland 2014-02-05 20:16:40
Incidentally, things are even cleaner if we take the logarithm base to be 5. Then we're fitting the sequence \[\log_52,2\log_52,3\log_52,\ldots,2014\log_52\] in between the terms of the sequence \[1,2,3\ldots,867.\]

copeland 2014-02-05 20:16:53
This is another place you might have seen this type of problem. If we let $\alpha=\log_52=\dfrac{\log2}{\log5}$ then we're now asking what the sequence $\lfloor \alpha n\rfloor$ looks like, since that's the left endpoint of the interval.

copeland 2014-02-05 20:17:00
The general result here is that since $\frac1\alpha$ is between 2 and 3, each integer appears either 2 or 3 times in the sequence. I strongly encourage you to try some examples later because these techniques are really powerful.

copeland 2014-02-05 20:18:10
Alright, everyone take 2 minutes to stretch and we'll start back in on the AMC12.

LMA 2014-02-05 20:20:21
will there be a transcript available for this?

copeland 2014-02-05 20:20:23
Yes. After class I will try to convince the transcripter to work. It should be up tonight.

DrMath 2014-02-05 20:20:47
the transcripter?

copeland 2014-02-05 20:20:52
It transcripts the transcripts.

copeland 2014-02-05 20:20:57
Seems like a natural name to me.

Bomist0 2014-02-05 20:21:51
should we start now?

copeland 2014-02-05 20:21:53
Yes we should.

copeland 2014-02-05 20:21:57
21. For every real number $x,$ let $\lfloor x \rfloor$ denote the greatest integer not exceeding $x,$ and let $$f(x) = \lfloor x \rfloor \left(2014^{x-\lfloor x \rfloor}-1\right).$$ The set of all numbers $x$ such that $1 \le x < 2014$ and $f(x) \le 1$ is a union of disjoint intervals. What is the sum of the lengths of those intervals?
$\phantom{peekaboo!}$
$\text{(A) } 1 \quad \text{(B) } \dfrac{\log{2015}}{\log{2014}} \quad \text{(C) } \dfrac{\log{2014}}{\log{2013}} \quad \text{(D) } \dfrac{2014}{2013} \quad \text{(E) } 2014^{\frac{1}{2014}}$

copeland 2014-02-05 20:22:06
Our function is $ f(x) = \lfloor x \rfloor \left(2014^{x-\lfloor x \rfloor}-1\right).$ Is there a way we can simplify this?

ninjataco 2014-02-05 20:22:57
$x-\lfloor x \rfloor = \{x\}$

lucylai 2014-02-05 20:22:57
let x=[x]+r

brian22 2014-02-05 20:22:57
[x]=x-floor(x)

Tuxianeer 2014-02-05 20:22:57
x-floor(x)={x}

AkshajK 2014-02-05 20:22:57

copeland 2014-02-05 20:23:06
The quantity $x-\lfloor x\rfloor$ is the "fractional part" of $x,$ which we usually write as $\{x\}.$ Now we want to talk about when \[ \lfloor x \rfloor (2014^{\{x\}}-1)\leq 1.\]

copeland 2014-02-05 20:23:23
What does the function $\lfloor x\rfloor$ look like?

dasobson 2014-02-05 20:24:00
lots of steps

dli00105 2014-02-05 20:24:00
a staircase

Tuxianeer 2014-02-05 20:24:00
discrete steps

swe1 2014-02-05 20:24:00
Staircase

dieButteristalle 2014-02-05 20:24:00
a staircase

alex31415 2014-02-05 20:24:00
Steps

CornSaltButter 2014-02-05 20:24:00
A staircase

DrMath 2014-02-05 20:24:00
it is a bunch of line segments

AaaDuhLoo 2014-02-05 20:24:00
Step function.

simon1221 2014-02-05 20:24:00
a set of stairs (graphicly)

checkmate1021 2014-02-05 20:24:00
a staircase without the thing on the side

AaaDuhLoo 2014-02-05 20:24:00
Staircase.

copeland 2014-02-05 20:24:05
(risers)

copeland 2014-02-05 20:24:23
The things on stairs that keep your change from going away and keep you from getting vertigo are called risers.

copeland 2014-02-05 20:24:29
It's a "step function":

copeland 2014-02-05 20:24:32

copeland 2014-02-05 20:24:35
What does the function $(2014^{\{x\}}-1)$ look like?

dasobson 2014-02-05 20:25:25
repeats over and over

brian22 2014-02-05 20:25:25
periodic

Tuxianeer 2014-02-05 20:25:25
repeated unit of a curve

lucylai 2014-02-05 20:25:25
it goes from 0 to 2013

alex31415 2014-02-05 20:25:25
Many exponential curves

DrMath 2014-02-05 20:25:25
exponential segments not exceeding 2013

anwang16 2014-02-05 20:25:25
diagonal repeats

blueferret 2014-02-05 20:25:25
repeated curves

fz0718 2014-02-05 20:25:25
a bunch of curves starting at the x axis

vincenthuang75025 2014-02-05 20:25:25
A weird curve thingy that keeps repeating

awesomemathlete 2014-02-05 20:25:25
the same exponential function repeated every integer

copeland 2014-02-05 20:25:27
The fractional part is periodic with period 1, so this function is periodic with period 1. In any period, it increases like a very sharp exponential, except it's shifted down by 1. When $x$ is an integer, the value is 0 and it approaches 2013 (a very big number) as $x$ increases to the next integer.

copeland 2014-02-05 20:25:31

copeland 2014-02-05 20:25:44
The product of the two functions will look like a bunch of copies of the exponential function scaled by different values.

copeland 2014-02-05 20:25:48

copeland 2014-02-05 20:26:20
Note: the curves start at 0 since $2014^0-1=1-1=0.$

copeland 2014-02-05 20:26:25
In function problems, it's often nice to have a picture of what's happening in mind. In this particular problem, the picture helps nail down what the intervals are. What are they?

copeland 2014-02-05 20:27:25
The intervals are the places where $f$ is less than 1.

copeland 2014-02-05 20:27:33
Can you tell me a place where $f$ is definitely less than 1?

Tuxianeer 2014-02-05 20:27:59
integers

lucylai 2014-02-05 20:27:59
integers

swe1 2014-02-05 20:27:59
At the integers

jweisblat14 2014-02-05 20:27:59
when x is an integer

coldsummer 2014-02-05 20:27:59
at integers

pedronr 2014-02-05 20:27:59
any integer

copeland 2014-02-05 20:28:08
$f$ is zero at all the integers, then $f$ increases.

copeland 2014-02-05 20:28:20
Therefore, what does our set of intervals look like?

swe1 2014-02-05 20:29:08
Intervals starting at the integers and get shorter

jweisblat14 2014-02-05 20:29:08
a little bit after each integer

blueferret 2014-02-05 20:29:08
they start from integers

Tuxianeer 2014-02-05 20:29:08
very small intervals starting at each integer

hametm 2014-02-05 20:29:08
From an integer to the point where the curve intersects y=1

ingridzhang97 2014-02-05 20:29:08
(0, something), (1, something) , (2, something), etc

checkmate1021 2014-02-05 20:29:08
a bunch of line segments starting at each integer that get smaller and smaller (exponentially?)

Dragon6point1 2014-02-05 20:29:08
A set of 2013 or so intervals starting at each integers and ending before the next.

jweisblat14 2014-02-05 20:29:08
a bunch of really short intervals, one startng at each integer and getting shorter and shorter

jiecut 2014-02-05 20:29:08
the intervals decrease

copeland 2014-02-05 20:29:17
For every integer 1 through 2013, there is one interval with left endpoint equal to that integer. The right endpoint occurs when the function $f$ hits the value 1.

copeland 2014-02-05 20:29:20
So we have 2013 intervals. The intervals are $[1,??]$ and $[2,??]$ and $[3,??]$ and $[4,??],$ etc. Now our goal is to figure out what the lengths of all these intervals are and add those up.

copeland 2014-02-05 20:29:30
Let's use casework on the left endpoints and then sum.

copeland 2014-02-05 20:29:32
The first left endpoint is 1. How wide is this interval?

alex31415 2014-02-05 20:30:22
log_2014(2)

lucylai 2014-02-05 20:30:22
log_2014(2)

joshxiong 2014-02-05 20:30:22
$\log_{2014} 2$

mathwizard888 2014-02-05 20:30:22

brian22 2014-02-05 20:30:22
log_2014(2)

Verjok 2014-02-05 20:30:22
log base 2014 of 2

ingridzhang97 2014-02-05 20:30:22
until 2014^{x}=2

copeland 2014-02-05 20:30:30
The right endpoint of the interval occurs when $f(x)=1.$ Therefore we want to solve \[ \lfloor x \rfloor (2014^{\{x\}}-1)=1\] when $\lfloor x\rfloor=1.$ That means we want to find the fractional part that solves \[2014^{\{x\}}-1=1.\]

copeland 2014-02-05 20:31:22
When we take the log of this equation we get \[\{x\}\log2014=\log2.\]

copeland 2014-02-05 20:31:28
The solution to this equation is \[\{x\}=\frac{\log 2}{\log{2014}}.\]

copeland 2014-02-05 20:31:35
Now I reflexively don't worry about the base of logs. It looks like a lot of you chose to take your logs with base 2014.

copeland 2014-02-05 20:32:11
That is also a fine approach, but only if you have the foresight to know that any logs we take in this problem are base 2014.

copeland 2014-02-05 20:32:29
I'm going to stick with the more general log for the moment, just to avoid getting in trouble.

copeland 2014-02-05 20:32:36
The first interval ends when $\lfloor x\rfloor=1$ and $\{x\}=\dfrac{\log2}{\log2014}.$ The interval is \[\left[1,1+\frac{\log2}{\log2014}\right]\] and the length is just $\dfrac{\log2}{\log2014}.$

copeland 2014-02-05 20:33:05
Let's make a chart that keeps track of the lengths of the intervals:
\[\begin{array}{c|l}
\lfloor x\rfloor&\text{length}\\
\hline
1&\frac{\log 2}{\log{2014}}
\end{array}\]

copeland 2014-02-05 20:33:08
The second left endpoint is 2. How wide is this interval?

jweisblat14 2014-02-05 20:33:46
the next one is log(3/2)/log(2014)

blueferret 2014-02-05 20:33:46
log2014(3/2)

lucylai 2014-02-05 20:33:46
log(3/2)/log(2014)

alex31415 2014-02-05 20:33:46
log(3/2)/log(2014)

Fibonacci97 2014-02-05 20:33:46
log(3/2)/log(2014)

DrMath 2014-02-05 20:33:46
log (3/2)/log(2014)

swe1 2014-02-05 20:33:46
log_2014(3/2)

hametm 2014-02-05 20:33:46
log(3/2)/log 2014

copeland 2014-02-05 20:33:49
The right endpoint is where $f(x)=1$ when $\lfloor x \rfloor = 2.$

copeland 2014-02-05 20:33:52
So we want to solve \[2(2014^{\{x\}}-1)=1.\] That means we want to find the fractional part that solves \[2014^{\{x\}}-1=\frac12.\]

copeland 2014-02-05 20:33:59
So $2014^{\{x\}} = \dfrac32,$ which means that \[\{x\}\log2014=\log\dfrac32.\]

copeland 2014-02-05 20:34:10
The solution to this equation is \[\{x\}=\frac{\log \frac32}{\log{2014}}.\]

copeland 2014-02-05 20:34:16
Can we make that prettier?

Pythonprogrammer 2014-02-05 20:34:48
log 3 - log2/ log 2014

mishai 2014-02-05 20:34:48
(log 3-log 2)/log 2014

lucylai 2014-02-05 20:34:48
(log(3)-log(2))/log(2014)

checkmate1021 2014-02-05 20:34:48
log(3) - log(2)

AaaDuhLoo 2014-02-05 20:34:48
(log3-log2)/(log2014) ?

mathbeida 2014-02-05 20:34:48
log3 - log 2

dasobson 2014-02-05 20:34:48
log 3 - log 2 over log 2014?

Fibonacci97 2014-02-05 20:34:48
log(3)-log(2)

copeland 2014-02-05 20:35:04
The solution to this equation simplifies as \[\{x\}=\frac{\log \frac32}{\log{2014}}=\frac{\log3-\log2}{\log2014}.\]

copeland 2014-02-05 20:35:08
The interval is \[\left[2,2+\frac{\log3-\log2}{\log2014}\right]\] and its length is $\dfrac{\log3-\log2}{\log2014}$

copeland 2014-02-05 20:35:11
Let's add this to the chart:\[\begin{array}{c|l}
\lfloor x\rfloor&\text{length}\\
\hline
1&\frac{\log 2}{\log{2014}}\\
2&\frac{\log3-\log2}{\log2014}\\
\end{array}\]

copeland 2014-02-05 20:35:14
The third left endpoint is 3. How wide is this interval?

Dragon6point1 2014-02-05 20:35:43
(log4 - log3)/log2014

anwang16 2014-02-05 20:35:43
(log4-log3)/log2014

joshualee2000 2014-02-05 20:35:43
log 4-log 3/log2014

tekgeek 2014-02-05 20:35:43
(log4-log3)/log2014

LightningX48 2014-02-05 20:35:43
(log4 - log3)/log(2014)

CornSaltButter 2014-02-05 20:35:43
${\frac{log4-log3}{log2014}}$

1915933 2014-02-05 20:35:48
(log 4 - log 3)/log 2014

bigc08 2014-02-05 20:35:48
log 4-log 3/log 2014

simon1221 2014-02-05 20:35:48
log4 - log3 / log201

copeland 2014-02-05 20:35:53
We need $f(x) = 1$ when $\lfloor x \rfloor = 3.$

copeland 2014-02-05 20:35:55
Now we want to solve \[3(2014^{\{x\}}-1)=1.\] That means we want to find the fractional part that solves \[2014^{\{x\}}-1=\frac13.\]

copeland 2014-02-05 20:36:01
So $2014^{\{x\}} = \dfrac43,$ and hence \[\{x\}\log2014 = \log\dfrac43=\log4-\log3.\]

copeland 2014-02-05 20:36:07
The solution to this equation is \[\{x\}=\frac{\log \frac43}{\log{2014}}=\frac{\log4-\log3}{\log2014}.\]

copeland 2014-02-05 20:36:10
\[\begin{array}{c|l}
\lfloor x\rfloor&\text{length}\\
\hline
1&\frac{\log 2}{\log{2014}}\\
2&\frac{\log3-\log2}{\log2014}\\
3&\frac{\log4-\log3}{\log2014}\\
\end{array}\]

copeland 2014-02-05 20:36:12
I bet you see the pattern now. What is the last term?

tekgeek 2014-02-05 20:37:14
+....+(log2014-log2013)/log2014

jweisblat14 2014-02-05 20:37:14
log 2014 - log 2013 / log 2014

Dragon6point1 2014-02-05 20:37:14
(log2014 - log2013)/log2014

DrMath 2014-02-05 20:37:14
(log 2014-log 2013)/log 2014

ingridzhang97 2014-02-05 20:37:14
(log2014-log2013)/log2014

mathnerd101 2014-02-05 20:37:14
(log2014-log2013)/log2014

LMA 2014-02-05 20:37:14
(log 2014-log2013)/log2014

DaChickenInc 2014-02-05 20:37:14
$\frac{\log 2014-\log 2013}{\log 2014}$

Tuxianeer 2014-02-05 20:37:14
(log(2014)-log(2013))/log(2014)

ferrastacie 2014-02-05 20:37:14
(log 2014 - log 2013)/ log 2014

ZekromReshiram 2014-02-05 20:37:14
(log2015-log2014)/log2014

copeland 2014-02-05 20:37:20
The final left endpoint is 2013.

copeland 2014-02-05 20:37:21
We need $f(x) = 1$ when $\lfloor x \rfloor = 2013.$

copeland 2014-02-05 20:37:24
Now we want to solve \[2013(2014^{\{x\}}-1)=1.\] That means we want to find the fractional part that solves \[2014^{\{x\}}-1=\frac1{2013}.\]

copeland 2014-02-05 20:37:29
This gives $2014^{\{x\}} = \dfrac{2014}{2013},$ so \[\{x\}\log2014= \log \dfrac{2014}{2013}=\log2014-\log2013.\]

copeland 2014-02-05 20:37:37
The solution to this equation is \[\{x\}=\frac{\log \frac{2014}{2013}}{\log{2014}}=\frac{\log2014-\log2013}{\log2014}.\]

copeland 2014-02-05 20:37:40
\[\begin{array}{c|l}
\lfloor x\rfloor&\text{length}\\
\hline
1&\frac{\log 2}{\log{2014}}\\
2&\frac{\log3-\log2}{\log2014}\\
3&\frac{\log4-\log3}{\log2014}\\
\vdots&~~~~~~~~\vdots\\
2013&\frac{\log2014-\log2013}{\log2014}\\
\end{array}\]

copeland 2014-02-05 20:37:48
What do we get when we add these?

want2learn 2014-02-05 20:38:27
1

ws0414 2014-02-05 20:38:27
A

hi how are you doing toda 2014-02-05 20:38:27
1

matholympiad25 2014-02-05 20:38:27
1 A

coldsummer 2014-02-05 20:38:27
1

ssk9208 2014-02-05 20:38:27
Cancel and cancel

mathtastic 2014-02-05 20:38:27
log2014/log2014=1

junat3 2014-02-05 20:38:27
A) 1

mathwizard888 2014-02-05 20:38:27
adding them gives log 2014/log 2014=1 (A)

doodlemaster7 2014-02-05 20:38:27
log 2014/2014

raylgheeheehee 2014-02-05 20:38:27
1

jiecut 2014-02-05 20:38:27
1

brainiac1 2014-02-05 20:38:27
1

Jayjayliu 2014-02-05 20:38:27
1

quackman 2014-02-05 20:38:27
1

IsabeltheCat 2014-02-05 20:38:27
1

tfmtoto 2014-02-05 20:38:27
$1$

jiecut 2014-02-05 20:38:27
log2014/log2014 = 1

flamefoxx99 2014-02-05 20:38:27
log2014/log2014

born_in_1999 2014-02-05 20:38:27
log2014/log2014

copeland 2014-02-05 20:38:32
When we add these the sum telescopes. The $\log2$ in row 1 cancels the $\log2$ in row 2. The $\log3$ in row 2 cancels the $\log 3$ in row 3. The only thing left is the final term\[\frac{\log2014}{\log2014}=\boxed1.\]The answer is (A).

copeland 2014-02-05 20:39:05
Incidentally, the first term ought to be $\dfrac{\log 2-\log1}{\log{2014}}.$

copeland 2014-02-05 20:39:32
Problem 12-22 was the same as 10-25.

copeland 2014-02-05 20:39:35
Nailed it.

simon1221 2014-02-05 20:39:53
3 left

copeland 2014-02-05 20:40:02
For anybody not keeping score.

copeland 2014-02-05 20:40:03
Let's keep going.

copeland 2014-02-05 20:40:05
23. The fraction $$\dfrac{1}{99^2} = 0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},$$ where $n$ is the length of the period of the repeating decimal expansion. What is the sum $b_0+b_1+\cdots + b_{n-1}?$
$\phantom{peekaboo!}$
$\text{(A) } 874 \quad \text{(B) } 883 \quad \text{(C) } 887 \quad \text{(D) } 891 \quad \text{(E) } 892$

copeland 2014-02-05 20:40:16
I love repeating decimal problems. I find them soothing.

copeland 2014-02-05 20:40:30
Know any well-known repeating decimals that we could possibly use here (assuming you don't already know the expansion of $1/99^2$)?

mathmaster2012 2014-02-05 20:41:01
1/99=0.010101010101

mssmath 2014-02-05 20:41:01
1/99

tapir1729 2014-02-05 20:41:01
1/99

distortedwalrus 2014-02-05 20:41:01
1/99 = .01010101...

mishai 2014-02-05 20:41:01
1/99 by itself

blueferret 2014-02-05 20:41:01
1/99=0.010101...

hametm 2014-02-05 20:41:01
1/99=.0101010101....

Dragon6point1 2014-02-05 20:41:01
$\frac{1}{99} = 0.01010101...$

maxplanck 2014-02-05 20:41:01
1/99

minimario 2014-02-05 20:41:01

vinayak-kumar 2014-02-05 20:41:01
1/99=.010101010101010101010101...

copeland 2014-02-05 20:41:04
You should definitely know that $\dfrac1{99}=.0101010101\cdots.$

copeland 2014-02-05 20:41:06
As a quick reminder because it's useful, we write \[1=.99999999\!\ldots.\] Now we are looking at a list of 99s:\[1=.\underbrace{99}\underbrace{99}\underbrace{99}\underbrace{99}\!\ldots.\]

copeland 2014-02-05 20:41:15
Once we see this, we can just divide each of these 99s by 99 to get 1:\[\frac1{99}=.\underbrace{01}\underbrace{01}\underbrace{01}\underbrace{01}\!\ldots.\]

copeland 2014-02-05 20:41:19
Now for the fun part. How do we get $\frac1{99^2}?$

copeland 2014-02-05 20:41:33
There are (at least) four nice ways to get at this. What do you see?

zoroark12345 2014-02-05 20:42:03
divide 99 again

DrMath 2014-02-05 20:42:03
divide by 99 again

bengals 2014-02-05 20:42:03
divide again

copeland 2014-02-05 20:42:13
There are multiple ways to divide, actually.

copeland 2014-02-05 20:42:15
We could use long division. This is a really nice approach and is often the best thing to turn to in a repeating decimals problem if you aren't struck with any other bit of inspiriation.

copeland 2014-02-05 20:42:25
We could also break the sum into a geometric series of powers of $10^{-2}$ as
\begin{align*}
\smash{\frac1{99}}=&\phantom{+ }.010000000000\!\ldots\\
&+ .000100000000\!\ldots\\
&+ .000001000000\!\ldots\\
&+ .000000010000\!\ldots\\
&+ .000000000100\!\ldots\\
&~~~~~~~~\vdots
\end{align*}
Then we could use our expression for $\dfrac1{99}$ to divide each of the terms by 99 and add.

copeland 2014-02-05 20:42:31
What else could we do?

DVA6102 2014-02-05 20:43:04
well 1/9801 = something/999....9999

bigc08 2014-02-05 20:43:04
divide each period in 0.0101010101 by 99

bigc08 2014-02-05 20:43:04
separate the decimal into periods

copeland 2014-02-05 20:43:10
We could figure out the period and write \[\frac{1}{99}=\frac{010101\!\ldots\!010101}{999999\!\ldots\!999999}\]and then just divide the numerator by 99. That definitely isn't a bad solution.

blueberry7 2014-02-05 20:43:47
square it!

blueferret 2014-02-05 20:43:47
square our number

swe1 2014-02-05 20:43:47
Squaring! Yaaayyy!

Tuxianeer 2014-02-05 20:43:47
square it

doodlemaster7 2014-02-05 20:43:47
square it!

yangwy 2014-02-05 20:43:47
1/99*1/99

awesomemathlete 2014-02-05 20:43:47
Square that decimal.

CornSaltButter 2014-02-05 20:43:47
square 1/99?

sirknightingfail 2014-02-05 20:43:47

copeland 2014-02-05 20:43:51
We could also just take the expression for $\dfrac1{99}$ and square it.

copeland 2014-02-05 20:43:56
Since the squaring solution is the most exotic-looking one here, I think we'll learn most from going with that one.

copeland 2014-02-05 20:44:00
\[\frac1{99^2}=(.0101010101010\!\ldots)^2.\]

copeland 2014-02-05 20:44:14
How does the square of this number start?

joshualee2000 2014-02-05 20:45:16
1/9801=0.00010203040506070809..........

jweisblat14 2014-02-05 20:45:16
.00010101010101...+.0000010101010101...+.000000010101010101...

dli00105 2014-02-05 20:45:16
0.00010203...

DigitalKing257 2014-02-05 20:45:16
0.0001020...

joshualee2000 2014-02-05 20:45:16
0.0001020304050607...

alex31415 2014-02-05 20:45:16
.00010203...

blueferret 2014-02-05 20:45:16
0.01020304...

sirknightingfail 2014-02-05 20:45:16

MrMerchant 2014-02-05 20:45:16
it turns out to be 0.001020304050607......

extremeaxe6 2014-02-05 20:45:16
0.00010203040506070809...

Pythonprogrammer 2014-02-05 20:45:16
000102030405...

jweisblat14 2014-02-05 20:45:16
.0001020304...

swe1 2014-02-05 20:45:16
0.000102030405?

copeland 2014-02-05 20:45:19
Long multiplication makes this easier.

copeland 2014-02-05 20:45:21
\[\begin{array}
\phantom{\times}.01010101010101\!\ldots\\
\times.01010101010101\!\ldots\\
\hline
\phantom{\times}.00010101010101010\!\ldots\\
+.00000101010101010\!\ldots\\
+.00000001010101010\!\ldots\\
+.00000000010101010\!\ldots\\
\phantom{abcdefg}\vdots\\
\hline
\phantom{+}.0001020304050607\!\ldots\\
\end{array}\]

copeland 2014-02-05 20:45:26
(Fun fact: this is actually the same sum you'd get if you divided the terms of the geometric series by 99 like in our other proposed solution.)

copeland 2014-02-05 20:45:31
That's intriguing, but I'm not sure what happens when we start carrying or how it ends or when it starts to repeat, so we need to think about those things.

copeland 2014-02-05 20:45:35
If we think of these as two-digit blocks, when does it start to carry from block to block?

jweisblat14 2014-02-05 20:46:26
99 to 100

lucylai 2014-02-05 20:46:26
at 9899...

superpi83 2014-02-05 20:46:26
at the 100th summand

brian22 2014-02-05 20:46:26
after ...9899

copeland 2014-02-05 20:46:28
We actually make it all the way up to 98 before anything weird happens.

copeland 2014-02-05 20:46:31
\[.000102030405\!\ldots\!969798?????\]

copeland 2014-02-05 20:46:37
Now the next block is 99 then 100 then 101, but that's definitely going to make a mess of things. Is there any way to fix this?

copeland 2014-02-05 20:46:59
Here's a hint: you almost always need to use $1=.99999\!\ldots$ somewhere in arguments like this.

swe1 2014-02-05 20:47:43
Subtract 0.999999999?

copeland 2014-02-05 20:47:46
Well that's too big.

copeland 2014-02-05 20:47:55
But is there a nice way to get a stream of nines out of this?

DrMath 2014-02-05 20:48:36
0.000......000999999.....

brainiac1 2014-02-05 20:48:36
subtract 0.00000.........9999999999

hametm 2014-02-05 20:48:36
Subtract .999999....*10^-9899

zoroark12345 2014-02-05 20:48:36
starting 99 subtract a bunch of 9's

Tuxianeer 2014-02-05 20:48:36
-0.000000000000(to lie with the 9899)999999999

copeland 2014-02-05 20:48:43
Let's pull out a 99 from all of the remaining terms. Then $99=99+0$ and $100=99+01$ and $101=99+02,$ etc. Now if we pull all the 99s out into a different series, we find something neat:

copeland 2014-02-05 20:48:44
\begin{align*}
&\phantom{+}.000102030405\!\ldots\!96979800010203\!\ldots\\
&+.000000000000\!\ldots\!00000099999999\!\ldots
\end{align*}

copeland 2014-02-05 20:48:51
Now that is the same as
\[\begin{array}
~&.000102030405\!\ldots\!96979800010203\!\ldots\\
+&.000000000000\!\ldots\!00000100000000\!\ldots\\
\hline
~&.000102030405\!\ldots\!96979900010203\!\ldots\\
\end{array}\]
and at this point, we could reasonably be convinced that we've found where the expansion starts to repeat.

copeland 2014-02-05 20:49:01
Let's prove that. We could figure out the period and divide things out, but now that we have a target, let's just see what happens when we multiply $x=.\overline{000102\!\ldots\!969799}$ by 99.

copeland 2014-02-05 20:49:09
Ugh. Multiplying by 99 is hard. What is easy?

ksun48 2014-02-05 20:49:33
*100 - 1

fcc1234 2014-02-05 20:49:33
Multiplying by 100-1

mathtastic 2014-02-05 20:49:33
multiplying by 100 then subtracting

CornSaltButter 2014-02-05 20:49:33
multiplying by 100, subtracting 1

tekgeek 2014-02-05 20:49:33
multiplying by 100-1

simon1221 2014-02-05 20:49:33
times 100, minus 1

mathtastic 2014-02-05 20:49:33
99x=100x-x

treemath 2014-02-05 20:49:33
multiplying by (100-1)

bookie331 2014-02-05 20:49:35
multiplying by 100 and subtract 1

copeland 2014-02-05 20:49:38
We could multiply $x$ by 100 and subtract $x$ because $99x=100x-x$.

copeland 2014-02-05 20:49:41
\begin{align*}
\phantom{-}100x=&.\overline{01020304\!\ldots\!96979900}\\
-\phantom{100}x=&.\overline{00010203\!\ldots\!95969799}\\
\hline
&.\overline{01010101\!\ldots\!01010101}
\end{align*}
At the end we get the really tricky $9900-9799=0101.$

copeland 2014-02-05 20:49:53
That is our expansion for $\dfrac1{99},$ so we're right!

copeland 2014-02-05 20:49:56
Now we need the sum of the digits of $.00010203\!\ldots\!969799.$ How can we get at this sum?

copeland 2014-02-05 20:50:18
WWGD?

doodlemaster7 2014-02-05 20:50:52
99*100/2 gauss

LMA 2014-02-05 20:50:52
does that stand for what would gauss do?

copeland 2014-02-05 20:50:57
Gauss, folks!

copeland 2014-02-05 20:51:05
How can we make the sum of these things easier?

bacca2002 2014-02-05 20:51:29
failed copeland

copeland 2014-02-05 20:51:33
Yeah, they can't all be winners.

checkmate1021 2014-02-05 20:52:02
sum in pairs

alex31415 2014-02-05 20:52:02
find the average

acegikmoqsuwy2000 2014-02-05 20:52:02
grouping them together?

extremeaxe6 2014-02-05 20:52:02
take 0+99, 1+98 etc and each summand adds to 99 or a digit sum of 18

copeland 2014-02-05 20:52:05
We're missing 98, but otherwise we have everything from 00 to 99.

copeland 2014-02-05 20:52:07
Let's pair the 2-digit numbers from 00 to 99 up and then subtract the contribution of 98.

copeland 2014-02-05 20:52:08
What do 00 and 99 contribute collectively to the sum?

mishai 2014-02-05 20:52:54
18

DrMath 2014-02-05 20:52:54
18

Pythonprogrammer 2014-02-05 20:52:54
18

jweisblat14 2014-02-05 20:52:54
18

dli00105 2014-02-05 20:52:54
18

distortedwalrus 2014-02-05 20:52:54
18

Dragon6point1 2014-02-05 20:52:54
18

lucylai 2014-02-05 20:52:54
18

ingridzhang97 2014-02-05 20:52:54
18

mssmath 2014-02-05 20:52:54
18

dli00105 2014-02-05 20:52:54
18

blueferret 2014-02-05 20:52:54
18

mathwizard888 2014-02-05 20:52:54
18

copeland 2014-02-05 20:52:57
00 and 99 contribute $0+0+9+9=18.$

copeland 2014-02-05 20:52:58
What do 47 and 52 contribute?

hametm 2014-02-05 20:53:17
18

ninjashiloh 2014-02-05 20:53:17
18

joshualee2000 2014-02-05 20:53:17
18

flamesofpi 2014-02-05 20:53:17
18

mathsd 2014-02-05 20:53:17
18

Chronix 2014-02-05 20:53:17
18

Weegee13 2014-02-05 20:53:17
18

hi how are you doing toda 2014-02-05 20:53:17
18

copeland 2014-02-05 20:53:20
47 and 52 contribute $4+7+5+2=18$ to the sum as well.

copeland 2014-02-05 20:53:21
What's the average sum of digits for a 2-digit number?

bigc08 2014-02-05 20:53:45
9

mxgo 2014-02-05 20:53:45
9

bacca2002 2014-02-05 20:53:45
9.

Dragon6point1 2014-02-05 20:53:45
9

jweisblat14 2014-02-05 20:53:45
9

treemath 2014-02-05 20:53:45
9

bengals 2014-02-05 20:53:45
9

math0127 2014-02-05 20:53:45
9

CyclicRain 2014-02-05 20:53:45
9

copeland 2014-02-05 20:53:47
The average sum of the digits of a 2-digit number is 9. How many total 2-digit numbers are there?

joshualee2000 2014-02-05 20:54:20
100

speck 2014-02-05 20:54:20
100

hametm 2014-02-05 20:54:20
100

WalkerTesla 2014-02-05 20:54:20
100

jiecut 2014-02-05 20:54:20
100

Tuxianeer 2014-02-05 20:54:20
100

copeland 2014-02-05 20:54:23
There are 100 total 2-digit numbers, so the sum of all the digits is 900.

copeland 2014-02-05 20:54:24
And the final answer?

jweisblat14 2014-02-05 20:55:00
883

lucylai 2014-02-05 20:55:00
883

CornSaltButter 2014-02-05 20:55:00
900-17=883

bacca2002 2014-02-05 20:55:00
since you have to subtract 9 and 8

pattycakechichi 2014-02-05 20:55:00
883 --> B

jweisblat14 2014-02-05 20:55:00
remove a 9+8-17 and get 883

howie2000 2014-02-05 20:55:00
$18\cdot50=900$, then $900-17=883$

WalkerTesla 2014-02-05 20:55:00
900-17 = 883

bacca2002 2014-02-05 20:55:00
so the final answer is (B) 883

guilt 2014-02-05 20:55:00
subtrat 17

abishek99 2014-02-05 20:55:00
900-17=883 B

howie2000 2014-02-05 20:55:00
$\boxed{883}$

harvey2014 2014-02-05 20:55:00
900-17=883

copeland 2014-02-05 20:55:02
We need to subtract the $9+8$ that we lose by skipping 98. That leaves $900-17=\boxed{883}$ as the sum. The answer is (B).

copeland 2014-02-05 20:55:52
(Incidentally, when I solved this one, I decided the average was something and I decided that the number of them was a multiple of 10, so the units digit had to be the same as the units digit of -9-8. There's only one answer here that ends in 3.)

pi37 2014-02-05 20:56:44
I did exactly that, except I subtracted 98 and chose the only answer that ended in 2

copeland 2014-02-05 20:56:51
So, there's a downside too.

copeland 2014-02-05 20:57:00
24. Let $f_0(x) = x + \left|x-100\right|-\left|x+100\right|,$ and for $n\ge 1,$ let $f_n(x) = \left|f_{n-1}(x)\right| - 1.$ For how many values of $x$ is $f_{100}(x)=0?$
$\phantom{peekaboo!}$
$\text{(A) } 299 \quad \text{(B) } 300 \quad \text{(C) } 301 \quad \text{(D) } 302 \quad \text{(E) } 303$

copeland 2014-02-05 20:57:08
What are some ways we can start in on this problem?

6stars 2014-02-05 20:58:05
We could graph?

mathtastic 2014-02-05 20:58:05
graphing it and trying values to get a feel for the function

ksun48 2014-02-05 20:58:05
and just try to solve for possible values of f_0(x)

DVA6102 2014-02-05 20:58:05
compute f_n for small n

joshualee2000 2014-02-05 20:58:05
find $f_1(x)$ ?

zoroark12345 2014-02-05 20:58:05
draw a graph

joshxiong 2014-02-05 20:58:05
Graph $f_0 (x)$

flyrain 2014-02-05 20:58:05
work with the recursive function

copeland 2014-02-05 20:58:10
There are generally two approaches to this problem - graphical and algebraic. I think we'll learn more from the graphical approach so that's the one I'm going to take.

copeland 2014-02-05 20:58:12
First let's graph the three components of $f_0.$ The first part is just $x$ which is easy to graph.

copeland 2014-02-05 20:58:16

copeland 2014-02-05 20:58:47
(To be honest, we'll graph for a while and probably switch to something more algebraic.)

copeland 2014-02-05 20:58:48
What does the graph of $y=|x-100|$ look like?

blueferret 2014-02-05 20:59:03
a v shape

Pythonprogrammer 2014-02-05 20:59:03
like a V!

dli00105 2014-02-05 20:59:03
aV s

copeland 2014-02-05 20:59:06
Where?

checkmate1021 2014-02-05 20:59:24
v centered at 100

blueferret 2014-02-05 20:59:24
the bottom is at x=100

fmasroor 2014-02-05 20:59:24
shifted -> 100

checkmate1021 2014-02-05 20:59:24
centered at 100

lee42 2014-02-05 20:59:24
A V shifted right by 100

LMA 2014-02-05 20:59:24
shifted right 100

mishai 2014-02-05 20:59:24
100 to the right of the origin

stonewater 2014-02-05 20:59:24
right 100

copeland 2014-02-05 20:59:26
It is an absolute value with slope $\pm1$ and vertex at 100.

copeland 2014-02-05 20:59:37
What about $y=-|x+100|?$

copeland 2014-02-05 20:59:38

trumpeter 2014-02-05 21:00:25
upside down v 100 left of the origin

DrMath 2014-02-05 21:00:25
similar, a minus V with vertex -100

az_phx_brandon_jiang 2014-02-05 21:00:25
upside down V at -100

bigc08 2014-02-05 21:00:25
flipped and vertex at -100,0

swe1 2014-02-05 21:00:25
Its the same except at -100 and flipped upside down

mathmaster2012 2014-02-05 21:00:25
same thing except base is on -100 and it is upside down

hamzacooly 2014-02-05 21:00:25
and upside down V centered at -100

mathbeida 2014-02-05 21:00:25
upside down v centered at -100

copeland 2014-02-05 21:00:28
This one is below the axis and has its vertex, a maximum, at -100.

copeland 2014-02-05 21:00:30

copeland 2014-02-05 21:00:39
Here are all three.

copeland 2014-02-05 21:00:42

copeland 2014-02-05 21:00:49
When we add these, we get a continuous, piecewise linear function. It has kinks at $x=\pm100.$ What are the values of $f_0$ at $\pm100?$

brian22 2014-02-05 21:01:43
-+100

joshualee2000 2014-02-05 21:01:43
+/-100

DrMath 2014-02-05 21:01:43
-100, 100

acegikmoqsuwy2000 2014-02-05 21:01:43
-100, and 100

blueferret 2014-02-05 21:01:43
-100 and +100

bigc08 2014-02-05 21:01:43
at 100 it is -100 and at -100 it is 100

WalkerTesla 2014-02-05 21:01:43
-100 and 100 for x = 100 and -100

speck 2014-02-05 21:01:43
-100 and 100

speck 2014-02-05 21:01:43
reverse the signs

ingridzhang97 2014-02-05 21:01:43
-100 and 100

copeland 2014-02-05 21:01:46
The values are
\begin{align*}
f_0(-100)&=-100+200-0=+100\\
f_0(100)&=100+0-200=-100.
\end{align*}

copeland 2014-02-05 21:01:52
Here is the graph.

copeland 2014-02-05 21:01:57

copeland 2014-02-05 21:01:59
Notice how there's always one slope that's different from the other two. That means the slope is always either $+1$ or $-1.$

copeland 2014-02-05 21:02:08
The graph rises from $-\infty$ to $(-100,100)$ then falls to $(100,-100)$ then rises again to $+\infty.$

copeland 2014-02-05 21:02:14

copeland 2014-02-05 21:02:25
We will solve the problem with a little algebra from here, but let's take a few moments to graph a few more of these. My original solution was entirely via graphs, but the algebraic ideas generalize a little better.

copeland 2014-02-05 21:02:35
Generalize to other problems, that is.

copeland 2014-02-05 21:02:37
Now how do we get $f_1?$

ws0414 2014-02-05 21:03:17
absolute, shift down one

dli00105 2014-02-05 21:03:17
absolute value minus 1

bacca2002 2014-02-05 21:03:17
take the absolute value and subtract 1

swe1 2014-02-05 21:03:17
Flip the part that is negative and shift down by one

pi37 2014-02-05 21:03:17
flip some stuff over and move it down

shandongboy 2014-02-05 21:03:17
turn all negatives into positive, shift down one unit

copeland 2014-02-05 21:03:20
We take the negative part of the graph and flip it. Then we move everything down 1.

copeland 2014-02-05 21:03:26

copeland 2014-02-05 21:03:31
For $f_2$ we do the same thing. We flip those little peaks and we subtract 1.

copeland 2014-02-05 21:03:33

copeland 2014-02-05 21:03:48
All of the values are shrinking to the $x$-axis and then the oscillate back and forth between -1 and 0.

copeland 2014-02-05 21:03:50
Alright, let's break from this and work purely algebraically.

copeland 2014-02-05 21:03:53
What should we do to determine if $f_{100}(x)=0?$

DrMath 2014-02-05 21:04:43
look at $f_{99}$?

Dragon6point1 2014-02-05 21:04:43
Find when $f_99 = 1$

blueferret 2014-02-05 21:04:43
f99(x)=1 or -1

1915933 2014-02-05 21:04:43
Determine if f_99=1 or -1.

DVA6102 2014-02-05 21:04:43
sorry f_99(x) = +-1

ws0414 2014-02-05 21:04:43

copeland 2014-02-05 21:04:48
If $f_{100}(x)=0$ then $f_{99}(x)$ must have been $-1$ or $+1.$

copeland 2014-02-05 21:04:51
Now what?

dli00105 2014-02-05 21:05:18
f98(x)

jweisblat14 2014-02-05 21:05:18
f98

copeland 2014-02-05 21:05:22
Good. What about it?

zacchro 2014-02-05 21:05:51
f98 is -2,0,2

Tuxianeer 2014-02-05 21:05:51
f_98=0,2,-2

joshxiong 2014-02-05 21:05:51
f_98(x) = 0, -2, 2

mathwizard888 2014-02-05 21:05:51
f98(x) is 0, -2, or 2

Tuxianeer 2014-02-05 21:05:51
f_98=0,2,-2 if f_100=0

awesomemathlete 2014-02-05 21:05:51
-2,0,2

LMA 2014-02-05 21:05:51
it must have been 0, 2, or -2?

joshualee2000 2014-02-05 21:05:51
then $f_(98)=0 or +/- 2

hametm 2014-02-05 21:05:51
It must have been 2, -2, or 0

copeland 2014-02-05 21:05:53
If $f_{99}(x)\in\{-1,+1\}$ then \[|f_{98}(x)|-1=-1\quad\quad\text{or}\quad\quad |f_{98}(x)|-1=+1.\]

copeland 2014-02-05 21:06:01
In the first case, $f_{98}(x)=0.$ In the second case $f_{98}(x)=\pm2.$ Therefore we must have $f_{98}(x)\in\{-2,0,2\}.$

copeland 2014-02-05 21:06:08
What does that tell us about $f_{97}(x)?$

ksun48 2014-02-05 21:06:50
3,1,-1,or negative thr3e

willwin4sure 2014-02-05 21:06:50
-3, -1, 1, 3

mathtastic 2014-02-05 21:06:50
-3 -1 1 3

ws0414 2014-02-05 21:06:50
then -2, -1, 0, 1, 2

DaChickenInc 2014-02-05 21:06:50
-3,-1,1,3

CornSaltButter 2014-02-05 21:06:50
{1,-1,3,-3}

copeland 2014-02-05 21:06:53
We know that $|f_{98}(x)|$ is one more than an element of $\{-2,0,2\}$ so has to be 1 or 3. That means $f_{98}(x)\in\{-3,-1,1,3\}.$

copeland 2014-02-05 21:06:55
The pattern is clear from here. $|f_{97}(x)|$ has to be one more than an element of $\{-3,-1,1,3\}$ so must be in $\{0,2,4\}$. That means $f_{97}(x)\in\{-4,-2,0,2,4\}.$

copeland 2014-02-05 21:06:57
And what set contains $f_0(x)?$

colinhy 2014-02-05 21:07:48
following the pattern gives us $f_0 \in {-100, -98, \ldots, 100}$

blueferret 2014-02-05 21:07:48
-100, -98, -96, ... +100

viva 2014-02-05 21:07:48
100, 98, ... 0, -98, ... -100

joshualee2000 2014-02-05 21:07:48
{-100,-98....98,100}

copeland 2014-02-05 21:07:54
We note that the index added to the largest element is constant, $100+0=99+1=98+2=97+3,$ etc. Therefore
\begin{align*}
f_{100}(x) = 0 &\Rightarrow f_{99}(x) \in \{-1,1\} \\
&\Rightarrow f_{98}(x) \in \{-2,0,2\} \\
&\Rightarrow f_{97}(x) \in \{-3,-1,1,3\} \\
&\vdots \\
&\Rightarrow f_0(x) \in \{-100,-98,-96,\ldots,96,98,100\}
\end{align*}

copeland 2014-02-05 21:08:07
Here's our graph again:

copeland 2014-02-05 21:08:12

copeland 2014-02-05 21:08:17
How many times is $f_0(x)=-100?$

joshualee2000 2014-02-05 21:08:42
2

Dragon6point1 2014-02-05 21:08:42
Twice.

blueferret 2014-02-05 21:08:42
2

WalkerTesla 2014-02-05 21:08:42
2

sirknightingfail 2014-02-05 21:08:42
2

simon1221 2014-02-05 21:08:42
2

zohee 2014-02-05 21:08:42
twice?

jkyman 2014-02-05 21:08:42
2

acegikmoqsuwy2000 2014-02-05 21:08:42
2

speck 2014-02-05 21:08:42
2

distortedwalrus 2014-02-05 21:08:45
2

roitishk 2014-02-05 21:08:45
2

copeland 2014-02-05 21:08:48
We noted that that is the minimum on the right. We hit the value 100 twice.

copeland 2014-02-05 21:08:49
How many times is $f_0(x)=-98?$

stonewater 2014-02-05 21:09:27
3

az_phx_brandon_jiang 2014-02-05 21:09:27
3

treemath 2014-02-05 21:09:27
3

checkmate1021 2014-02-05 21:09:27
3

speck 2014-02-05 21:09:27
3

guilt 2014-02-05 21:09:27
3

ingridzhang97 2014-02-05 21:09:27
3

Jayjayliu 2014-02-05 21:09:27
3

mikep 2014-02-05 21:09:27
3

lee42 2014-02-05 21:09:27
thrice?

copeland 2014-02-05 21:09:30
We're between the peak and the trough so we hit this value three times. What else?

ksun48 2014-02-05 21:10:15
here are the possible values 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2

blueferret 2014-02-05 21:10:15
all of them hit 3 times except -100 and 100

Tuxianeer 2014-02-05 21:10:15
3 for everything -98 to 98

Dragon6point1 2014-02-05 21:10:15
We hit all other values except $100$ three times.

speck 2014-02-05 21:10:15
100 and -100 are 2, the rest are 3

mathtastic 2014-02-05 21:10:15
you wtill hit everything else 3 times as well, but 100 you will only hit twice

swe1 2014-02-05 21:10:15
Every number from -98 to 98 intersects three times

stonewater 2014-02-05 21:10:15
it's always 3 until +100, which is 2

zohee 2014-02-05 21:10:15
It'll be three for everything except for the two extreme values

copeland 2014-02-05 21:10:18
We hit all the values three times except $f_0(x)=-100$ twice and $f_0(x)=100$ twice.

copeland 2014-02-05 21:10:19
How many total solutions are there?

bacca2002 2014-02-05 21:10:51
so the answer is just 3*99+4

bacca2002 2014-02-05 21:10:51
301

bacca2002 2014-02-05 21:10:51
99*3+4

bacca2002 2014-02-05 21:10:51
answer is C.

WalkerTesla 2014-02-05 21:10:51
99*3+4 = 301

1915933 2014-02-05 21:10:51
(c) 301

dantheman567 2014-02-05 21:10:51
301

blueferret 2014-02-05 21:10:51
3*99+2*2

speck 2014-02-05 21:10:51
2*2+3*99=301

sirknightingfail 2014-02-05 21:10:51
3*99+2*2=301

DrMath 2014-02-05 21:10:51
303-2=301C

Pythonprogrammer 2014-02-05 21:10:51
(C) 301

copeland 2014-02-05 21:10:53
We have 101 total numbers on the list. We hit all of them three times except the first and last. Therefore the total is $99\cdot3+2\cdot2=301.$ The answer is (C).

copeland 2014-02-05 21:10:59
Hm.

copeland 2014-02-05 21:11:04
Does anybody know what time it is?

CornSaltButter 2014-02-05 21:11:25
The big one!!!!

brainiac1 2014-02-05 21:11:25
YAY One laft

zohee 2014-02-05 21:11:25
Final problem time?

DVA6102 2014-02-05 21:11:25
time for number 25

CornSaltButter 2014-02-05 21:11:25
adventure time

copeland 2014-02-05 21:11:42
Yes. We're at 12-25. Let's have at it.

copeland 2014-02-05 21:11:43
25. The parabola $P$ has focus $(0,0)$ and goes through the points $(4,3)$ and $(-4,-3).$ For how many points $(x,y) \in P$ with integer coordinates is it true that $|4x+3y| \le 1000?$
$\phantom{peekaboo!}$
$\text{(A) } 38 \quad \text{(B) } 40 \quad \text{(C) } 42 \quad \text{(D) } 44 \quad \text{(E) } 46$

copeland 2014-02-05 21:12:18
Alright, we're given the focus $(0,0)$ and two points $(4,3)$ and $(-4,-3).$ Usually how many points do you need to describe a conic section?

swe1 2014-02-05 21:12:50
5

DrMath 2014-02-05 21:12:50
5

brandbest1 2014-02-05 21:12:50
5

distortedwalrus 2014-02-05 21:12:50
5

quackman 2014-02-05 21:12:50
5

dantheman567 2014-02-05 21:12:50
FIVE!!!

copeland 2014-02-05 21:12:53
Usually you need 5 points to define a conic section. Well, at least when the points are on the conic.

copeland 2014-02-05 21:12:55
That's what jumped out at me at first when I saw this problem. Something weird is definitely going on here. Let's just say we have no idea how many parabolas fit this description.

copeland 2014-02-05 21:13:00
Incidentally, there are two things that we missed in our "three points" analysis. First is that we know it's a parabola: any 4 well-chosen points define exactly 2 parabolas.

copeland 2014-02-05 21:13:11
Second, we're not just given random points - we're given a focus. It turns out that foci are just magicially twice as good as other points for defining a conic.

copeland 2014-02-05 21:13:19
Now, none of what we just said is going to help us too much on this problem, but if you have all these ideas on-hand you can look at this problem and say to yourself, "Self, there are two parabolas here. That's not too scary and might come in handy. At least we need to hope that both parabolas give the same answer."

copeland 2014-02-05 21:13:34
Incidentally, I assume you don't know any of that stuff - it's a shame that nobody really teaches conics. It's really beautiful mathematics.

copeland 2014-02-05 21:13:39
Let's get started.

copeland 2014-02-05 21:13:49
What is the focus of a parabola? What is the pertinent definition for a parabola?

blueferret 2014-02-05 21:14:51
identical distance between focus and a line, the directrix

dasobson 2014-02-05 21:14:51
equidistant from focus and directrix

LMA 2014-02-05 21:14:51
the set of points equidistant from the focus and the directrix

ws0414 2014-02-05 21:14:51
set of points equidistant from a point and a line

djmathman 2014-02-05 21:14:51
A parabola is the locus of points that are equidistant from the focus and a line called the directrix

joshxiong 2014-02-05 21:14:51
A parabola is the set of points that are equidistant to a point (focus) and a line (directrix)

alex31415 2014-02-05 21:14:51
Distance from any point to the focus is the same as distance to directrix

richard4912 2014-02-05 21:14:51
Distance from any point on the parabola is equidistant from the focus and the directrix

math_posse 2014-02-05 21:14:51
Distance from focus and directrix is equal

hexagram 2014-02-05 21:14:51
same distance from directrix to focus

copeland 2014-02-05 21:14:54
Given a point (the focus) and a line (the directrix) we can define a parabola as the locus of all points equidistant from the focus and the directrix.

copeland 2014-02-05 21:15:06
I'm impressed.

copeland 2014-02-05 21:15:08
What do you notice about our three points?

MSTang 2014-02-05 21:15:49
symmetric about origin

guilt 2014-02-05 21:15:49
collinear

dli00105 2014-02-05 21:15:49
collinear

mathwizard888 2014-02-05 21:15:49
collinear

DaChickenInc 2014-02-05 21:15:49
collinear

checkmate1021 2014-02-05 21:15:49
colinear

mathnerd101 2014-02-05 21:15:49
the 2 points are 5 units away from the focus

swe1 2014-02-05 21:15:49
They are symmetrical around the origin

checkmate1021 2014-02-05 21:15:49
collinear

willwin4sure 2014-02-05 21:15:49
all on the same line

hametm 2014-02-05 21:15:49
Origen and two points that have rotational symmetry

mssmath 2014-02-05 21:15:49
the midpoint is the origin

acegikmoqsuwy2000 2014-02-05 21:15:49
they are reflected about the origin?

copeland 2014-02-05 21:15:52
These three points are collinear. I bet that's going to make things easier.

copeland 2014-02-05 21:15:54
The focus is also the midpoint between the other two.

copeland 2014-02-05 21:15:56
What does that make the segment connecting $A=(4,3)$ to $B=(-4,-3)?$

ws0414 2014-02-05 21:16:19
the something...

copeland 2014-02-05 21:16:20
Exactly!

swe1 2014-02-05 21:16:33
latus rectum

joshxiong 2014-02-05 21:16:33
latus rectum

djmathman 2014-02-05 21:16:33
the latus rectum of the parabola

ingridzhang97 2014-02-05 21:16:33
the latus something

storm7 2014-02-05 21:16:33
latus rectum

copeland 2014-02-05 21:16:37
That's the latus rectum of the parabola. A chord in a conic that is bisected by the focus is called a latus rectum. In a circle, diameters are latus recti. In a hyperbola or ellipse, there are two of these - one for each focus.

copeland 2014-02-05 21:16:46
What do you know about the latus rectum that will probably be useful?

Tuxianeer 2014-02-05 21:17:25
parallel to the directrix

ingridzhang97 2014-02-05 21:17:25
parallel to the directrix

djmathman 2014-02-05 21:17:25
it's parallel to the directrix

shandongboy 2014-02-05 21:17:25
parallel to directrix?

DaChickenInc 2014-02-05 21:17:25
parallel to directrix?

mathwizard888 2014-02-05 21:17:25
parallel to the directrix

dantheman567 2014-02-05 21:17:25
parallel to the conic section directrix

1915933 2014-02-05 21:17:25
The latus rectum is always parallel to the directix.

copeland 2014-02-05 21:17:28
The latus rectum is parallel to the directrix. Therefore we know the slope of the directrix. The latus rectum is also perpendicular to the axis of symmetry. Now we're on our way to being able to draw a really nice picture.

copeland 2014-02-05 21:17:44
What else?

bradleyramunas 2014-02-05 21:18:15
1/4p

dasobson 2014-02-05 21:18:15
equal to |4a|

DrMath 2014-02-05 21:18:15
its length is 4a or something like that

trumpeter 2014-02-05 21:18:15
four times distance from focus to vertex

ws0414 2014-02-05 21:18:15
four times distance from focus to center of parabola

mikep 2014-02-05 21:18:15
its length is 4p

joshualee2000 2014-02-05 21:18:15
4a?

copeland 2014-02-05 21:18:28
That's good. We will draw a picture to get that in a second.

copeland 2014-02-05 21:18:34
What about the axis of symmetry?

Modest_Ked 2014-02-05 21:19:25
perpendicular

dasobson 2014-02-05 21:19:25
perpendicular and goes through the origin

bacca2002 2014-02-05 21:19:25
it's perpendicular to the directrizx

CornSaltButter 2014-02-05 21:19:25
it's perpendicular to the axis of symmetry

LMA 2014-02-05 21:19:25
it's perpindicular to the directrix/latus rectum

ingridzhang97 2014-02-05 21:19:25
perpendicular to the latus rectum

yvetal27 2014-02-05 21:19:25
the axis is perpendicular to the directx

jweisblat14 2014-02-05 21:19:25
perpendicular to the latus rectum, through the focus

copeland 2014-02-05 21:19:27
The axis of symmetry ought to be perpendicular to our segment.

copeland 2014-02-05 21:19:37

copeland 2014-02-05 21:19:40
Here's our latus rectum.

copeland 2014-02-05 21:19:43
How far is $A=(4,3)$ from the directrix?

mathtastic 2014-02-05 21:20:04
5

DaChickenInc 2014-02-05 21:20:04
5

DrMath 2014-02-05 21:20:04
5

bacca2002 2014-02-05 21:20:04
5.

ksun48 2014-02-05 21:20:04
5

Allan Z 2014-02-05 21:20:04
5

viva 2014-02-05 21:20:04
5

copeland 2014-02-05 21:20:07
The point $A$ is 5 units from the focus $(0,0)$ so it is 5 units from the directrix.

copeland 2014-02-05 21:20:13
How far is the focus from the directrix?

dantheman567 2014-02-05 21:20:41
also 5?

mathtastic 2014-02-05 21:20:41
again 5

awesomemathlete 2014-02-05 21:20:41
5

blueferret 2014-02-05 21:20:41
5

legozelda 2014-02-05 21:20:41
5

DVA6102 2014-02-05 21:20:41
5

copeland 2014-02-05 21:20:46
Since the directrix is parallel to the latus rectum and the line of symmetry of the diagram is perpendicular to the latus rectum, when we drop the altitudes from $A$ and $B$ to the directrix, we get a rectangle.

copeland 2014-02-05 21:20:55

copeland 2014-02-05 21:20:58
This is a $5\times 10$ rectangle. The distance from the focus to the directrix is the height of this rectangle, 5. Now we can draw our parabola:

copeland 2014-02-05 21:21:05

copeland 2014-02-05 21:21:16
Notice that here is where we see our two parabolas. The directrix could be located either direction from the latus rectum.

copeland 2014-02-05 21:21:21

copeland 2014-02-05 21:21:24
Let's just hope that the answer is independent of this choice.

copeland 2014-02-05 21:21:26
First off, what is bad about this parabola?

blueferret 2014-02-05 21:22:13
its slanted

dantheman567 2014-02-05 21:22:13
its slanted

TheUnChosenOne 2014-02-05 21:22:13
tilted

Dragon6point1 2014-02-05 21:22:13
It's tilted.

checkmate1021 2014-02-05 21:22:13
it is tilted

ScottBusche 2014-02-05 21:22:13
It's slanted.

hametm 2014-02-05 21:22:13
It's axis of symmetry isn't parallel to an axis

flyrain 2014-02-05 21:22:13
it's not vertical or horizontal it is not a function

copeland 2014-02-05 21:22:19
It's all wonky. It would be way better if it weren't. How can we fix it?

acegikmoqsuwy2000 2014-02-05 21:22:54
rotate it

joshualee2000 2014-02-05 21:22:54
turn it

flyrain 2014-02-05 21:22:54
rotate the axis

Pythonprogrammer 2014-02-05 21:22:54
rotation

blueferret 2014-02-05 21:22:54
tilt the axes?

quackman 2014-02-05 21:22:54
rotation

hametm 2014-02-05 21:22:54
Tilt the axis?

distortedwalrus 2014-02-05 21:22:54
introduce a new coordinate system

dli00105 2014-02-05 21:22:54
rotate it

sirknightingfail 2014-02-05 21:22:54
rotate your axis

copeland 2014-02-05 21:23:08
We could rotate the parabola or rotate the axes.

copeland 2014-02-05 21:23:34
It's morally the same. Since we actually care about $x$ and $y,$ let's introduce new coordinates and leave the parabola alone.

copeland 2014-02-05 21:23:46
Let the $u$-axis go through $A$ and $B$ and let the $v$-axis be perpendicular to that through the focus.

copeland 2014-02-05 21:23:51

copeland 2014-02-05 21:24:15
I want to come clean with you here. I'm morally obligated to pretend that this problem can be solved in a reasonable amount of time without using any linear algebra. I'll stick to that narrative, but I don't really believe it's possible. Anyway, let's keep going.

copeland 2014-02-05 21:24:29
What is the equation for the $u$-axis?

joshxiong 2014-02-05 21:25:15
y = 3/4 x

CornSaltButter 2014-02-05 21:25:15
y= 3x/4

blueferret 2014-02-05 21:25:15
y=3/4x

simon1221 2014-02-05 21:25:15
y=-3/4x

distortedwalrus 2014-02-05 21:25:15
y=3/4x

stonewater 2014-02-05 21:25:15
y=3/4x

mathwizard888 2014-02-05 21:25:15
4y=3x

math_posse 2014-02-05 21:25:15
y=3/4x

copeland 2014-02-05 21:25:18
The $u$-axis has slope of $\frac34$ and goes through the origin so its equation is $y=\frac34x.$ That looks better in standard form as $3x-4y=0.$

copeland 2014-02-05 21:25:23
What is the equation for the $v$-axis?

storm7 2014-02-05 21:25:58
y=-4/3x

dasobson 2014-02-05 21:25:58
y=-4x/3

Modest_Ked 2014-02-05 21:25:58
y=-4/3x

hametm 2014-02-05 21:25:58
4x+3y=0

blueferret 2014-02-05 21:25:58
3y+4x=0

az_phx_brandon_jiang 2014-02-05 21:25:58
4x+3y=0

joshualee2000 2014-02-05 21:25:58
4x+3y=0

stonewater 2014-02-05 21:25:58
4x+3y=0

storm7 2014-02-05 21:25:58
4x+3y=0

LMA 2014-02-05 21:25:58
y=-4/3x

bradleyramunas 2014-02-05 21:25:58
y=-4/3x

copeland 2014-02-05 21:26:00
The $v$-axis is perpendicular to the $u$-axis. It has slope $-\frac43$ and goes through the origin. Its equation is $4x+3y=0.$

copeland 2014-02-05 21:26:10
Incidentally, why is standard form useful in a problem like this?

ingridzhang97 2014-02-05 21:27:08
to find distance between point and line and such

qwertyu 2014-02-05 21:27:08
Can easily find distance

joshxiong 2014-02-05 21:27:08
distance formula for a point and a line

copeland 2014-02-05 21:27:11
The distance formula from a point to a line needs the standard form.

copeland 2014-02-05 21:27:13
Now the $u$-coordinate is the signed distance from the $v$-axis. What is the $(u,v)$ coordinate for a point with coordinate $(x,y)$ in our original frame?

copeland 2014-02-05 21:28:16
We need the distance from a point to a line. (Hopefully you remember the formula for the distance from a point to a line!)

copeland 2014-02-05 21:28:39
What is the distance from $(x,y)$ to the line $4x+3y=0?$

mathbeida 2014-02-05 21:29:22
|4x+3y|/5

blueferret 2014-02-05 21:29:22
(4x+3y)/5

DrMath 2014-02-05 21:29:22
|4x+3y|/5

zhuangzhuang 2014-02-05 21:29:22
|4x+3y|/5

mikep 2014-02-05 21:29:22
abs(4x+3y)/5

ingridzhang97 2014-02-05 21:29:22
|4x+3y|/sqrt(4^2+3^2)

1915933 2014-02-05 21:29:22
(4x+3y)/5

mishai 2014-02-05 21:29:22
|4x+3y|/5

copeland 2014-02-05 21:29:26
The distance from $(x,y)$ to $4x+3y=0$ is \[\frac{|4x+3y|}{\sqrt{3^2+4^2}}.\]Since we want a signed distance, we drop the absolute values and just get $u=\frac{4x+3y}5.$ This is positive in the first $(x,y)$-quadrant.

copeland 2014-02-05 21:29:34
Likewise, the $v$-coordinate is the distance from $(x,y)$ to the $u$-axis, \[\frac{-3x+4y}5.\]I picked the sign here that is positive in the second $(x,y)$-quadrant.

copeland 2014-02-05 21:29:44
\begin{align*}
u&=\frac{4x+3y}5,\\
v&=\frac{-3x+4y}5.
\end{align*}

copeland 2014-02-05 21:29:53
Look back at the problem. What tells us we're on the right track?

DVA6102 2014-02-05 21:30:42
the equation is |4x+3y|<=1000

zhuangzhuang 2014-02-05 21:30:42
the desired involves |4x+3y|

ws0414 2014-02-05 21:30:42
the 4x+3y

infiniteturtle 2014-02-05 21:30:42
the $4x+3y$ part

jweisblat14 2014-02-05 21:30:42
4x + 3y

acegikmoqsuwy2000 2014-02-05 21:30:42
we have the |4x+3y| part

mathnerd101 2014-02-05 21:30:42
Theres a 4x+3y in the question?

dli00105 2014-02-05 21:30:42
5|u|<=1000

copeland 2014-02-05 21:30:44
One of our constraints is $|4x+3y|\leq1000.$ That is a multiple of $u.$ We can divide by 5 and rewrite the constraint as

copeland 2014-02-05 21:30:45
$|u|\leq200.$

copeland 2014-02-05 21:31:08
We just want the lattice points between these two lines, $u=-200$ and $u=200.$

copeland 2014-02-05 21:31:12
Alright, good. Now let's get an equation for the parabola in our $u,v$-coordinates. We have the distance definition for the parabola. What is the distance from the $(u,v)$ to the focus?

mishai 2014-02-05 21:32:02
sqrt(u^2+v^2)

brainiac1 2014-02-05 21:32:02
sqrt(u^2+v^2)

joshualee2000 2014-02-05 21:32:02
$\sqrt(u^2+v^2)$

AkshajK 2014-02-05 21:32:02

copeland 2014-02-05 21:32:07
The distance is just $\sqrt{u^2+v^2}.$ (Cool computation: check later that $\sqrt{u^2+v^2}=\sqrt{x^2+y^2}.$ It's neat and helps to clarify the coordinates some.)

copeland 2014-02-05 21:32:12
Now we need the directrix. What do we know about it?

copeland 2014-02-05 21:32:26
Specifically we need its equation.

zhuangzhuang 2014-02-05 21:33:01
parallel to axis

ingridzhang97 2014-02-05 21:33:01
parallel to the u-axis

LMA 2014-02-05 21:33:01
it's parallel to u

1915933 2014-02-05 21:33:01
It is parallel to the u-axis.

ingridzhang97 2014-02-05 21:33:01
5 units below the u axis

LMA 2014-02-05 21:33:01
v=-5

hametm 2014-02-05 21:33:01
v=-5

joshualee2000 2014-02-05 21:33:01
its equation is v=-5

brainiac1 2014-02-05 21:33:01
it's 5 below the latus rectum

DaChickenInc 2014-02-05 21:33:01
$v=-5$

storm7 2014-02-05 21:33:01
v=-5

copeland 2014-02-05 21:33:04
We know it's parallel to the $u$-axis, so its equation is $v=\text{something}.$ We also know it is a distance of 5 from the focus, so it must be $v=5$ or $v=-5.$ Let's pick $v=-5$ since that's the one we drew above.

copeland 2014-02-05 21:33:05
What is the distance from $(u,v)$ to the directrix?

blueferret 2014-02-05 21:34:42
v+5

mathwizard888 2014-02-05 21:34:42
|v+5|

DaChickenInc 2014-02-05 21:34:42
$|v+5|$

jweisblat14 2014-02-05 21:34:42
v+5

joshualee2000 2014-02-05 21:34:48
v+5

copeland 2014-02-05 21:34:50
The distance to the directrix is just $|v+5|$.

copeland 2014-02-05 21:34:53
Therefore the equation for the parabola is \[\sqrt{u^2+v^2}=|v+5|.\]

copeland 2014-02-05 21:34:56
That's ugly. . .

DrMath 2014-02-05 21:35:14
square

awesomemath123 2014-02-05 21:35:14
square it

dli00105 2014-02-05 21:35:14
square it

WalkerTesla 2014-02-05 21:35:14
Square both sides

1915933 2014-02-05 21:35:14
Square both sides!

prime235 2014-02-05 21:35:14
square both sides

copeland 2014-02-05 21:35:18
If we square it we get \[u^2+v^2=(v+5)^2.\]

copeland 2014-02-05 21:35:18
What happens when we simplify this?

zhuangzhuang 2014-02-05 21:35:59
u^2=10v+25

Tuxianeer 2014-02-05 21:35:59
u^2=10v+25

mathtastic 2014-02-05 21:35:59
v^2 cancel

dasobson 2014-02-05 21:35:59
u^2-10v-25=0

mathtastic 2014-02-05 21:35:59
u^2=10v+25

nowunkie 2014-02-05 21:35:59
u^2+v^2=v^2+10v+25

maxplanck 2014-02-05 21:35:59
u^2 = 10v + 25

prime235 2014-02-05 21:35:59
v^2 disappears!

CornSaltButter 2014-02-05 21:35:59
${u^2-10v-25=0}$

nowunkie 2014-02-05 21:35:59
u^2=10v+25

AkshajK 2014-02-05 21:35:59

mikeli380 2014-02-05 21:35:59
$u^2=10v+25

checkmate1021 2014-02-05 21:35:59
$u^2 -10v-25 = 0$

copeland 2014-02-05 21:36:03
First we get \[u^2+v^2=v^2+10v+25\]and then the $v^2$ terms cancel giving us the parabola

copeland 2014-02-05 21:36:04
\[u^2=10v+25.\]

copeland 2014-02-05 21:36:11
Gosh, that's pretty. That means we have the right coordinate system.

copeland 2014-02-05 21:36:14
Usually a parabola would have the squares of both coordinates. It's only when you pick the axis of symmetry as one of your coordinates that one of the squares drops away.

copeland 2014-02-05 21:36:39
"to be parallel to one of the coordinate axes," I should have said.

copeland 2014-02-05 21:36:41
Now it is time to go looking for our integer solutions. Remember, though, that $u$ and $v$ don't have to be integers. Just $x$ and $y.$ What should we do?

blueferret 2014-02-05 21:37:27
substitute them back in

dasobson 2014-02-05 21:37:27
write in terms of x and y maybe?

mssmath 2014-02-05 21:37:27
convert back

ws0414 2014-02-05 21:37:27
convert back

hametm 2014-02-05 21:37:27
Plug in x and y

mishai 2014-02-05 21:37:27
u=(4x+3y)/5

checkmate1021 2014-02-05 21:37:27
use what we found for u and v in terms of x and y

copeland 2014-02-05 21:37:30
Let's rewrite the equation in terms of $x$ and $y$. That gives\[\left(\frac{4x+3y}5\right)^2=10\left(\frac{-3x+4y}5\right)+25.\]

copeland 2014-02-05 21:37:34
What do you notice?

copeland 2014-02-05 21:38:19
Specifically, we have some diophantine thing here.

copeland 2014-02-05 21:38:34
We're looking for our integer solutions so we need to put our number theory hats on.

distortedwalrus 2014-02-05 21:38:53
I don't own one of those.

copeland 2014-02-05 21:38:56
Borrow one?

copeland 2014-02-05 21:39:22
We want to talk about the integer solutions in $x$ and $y$ now. Say something that is like, "something something integer something."

DVA6102 2014-02-05 21:40:27
because 5|10, RHS must be an integer, so LHS must be an integer as well

copeland 2014-02-05 21:40:46
OK, what?

copeland 2014-02-05 21:41:03
What happens when we start plugging integers in for $x$ and $y?$

blueferret 2014-02-05 21:41:30
both sides are integers!

Allan Z 2014-02-05 21:41:30
5|(4x+3y)

ilikepie333 2014-02-05 21:41:30
4x+3y is divisible by 5

mishai 2014-02-05 21:41:30
4x+3y must be divisible by 5

ingridzhang97 2014-02-05 21:41:30
4x+3y must be divisible by 5

copeland 2014-02-05 21:41:46
$v$ is not an integer. It has a 5 in the denominator.

copeland 2014-02-05 21:41:54
However, $10v$ is an integer. The right side of the equation is always an integer so the left side must be. That forces $u$ to be an integer and specifically we know that \[4x+3y\equiv0\pmod5.\]

copeland 2014-02-05 21:42:21
There's a chance that says something about $v$ also. Can we learn anything about $-3x+4y$ now?

bob12345678 2014-02-05 21:42:59
its also 0 in mod 5

dli00105 2014-02-05 21:42:59
-3x+4y=0(mod5)

mathtastic 2014-02-05 21:42:59
yes it must also be congruent to 0 mod 5!!!

tim9099xxzz 2014-02-05 21:42:59
Also divisble by 5

copeland 2014-02-05 21:43:02
Why?

AkshajK 2014-02-05 21:43:29
multiply by 3 and take mod 5

nowunkie 2014-02-05 21:43:29
4y-3x has to be divisible by 5

CornSaltButter 2014-02-05 21:43:29
2(-3x+4y), so 5 | 4y-3x

copeland 2014-02-05 21:43:36
If we multiply \[4x+3y\equiv0\pmod5\] by 3 we get \[12x+9y\equiv0\pmod5.\] That is equivalent to \[-3x+4y\equiv0\pmod5.\]

copeland 2014-02-05 21:43:55
There are other ways to get the same result. Like:

DaChickenInc 2014-02-05 21:43:57
$x\equiv3y\pmod{5}$

copeland 2014-02-05 21:44:04
And then grind through.

copeland 2014-02-05 21:44:13
Specifically, though, we just learned that $v$ is also an integer. Awesome.

copeland 2014-02-05 21:44:20
Let's go back to the parabola,\[u^2=10v+25.\] Now we know $u$ and $v$ are integers. However we notice that the right side is now divisible by 5. That makes the left side divisible by 5 and $u$ is a multiple of 5.

copeland 2014-02-05 21:44:31
This forces $u$ to be on the list $-200,-195,-190,\ldots,190,195,200.$ There are still 81 numbers in the list and that's not one of the possible answer choices so we need to keep going.

copeland 2014-02-05 21:44:37
What have we missed?

copeland 2014-02-05 21:45:41
Here's a hint; the answers are all around 40. . . .

mathwizard888 2014-02-05 21:46:07
multiples of 10 don't work

zhuangzhuang 2014-02-05 21:46:07
parity of v

ABCDE 2014-02-05 21:46:07
odd multiple of 5

pi37 2014-02-05 21:46:07
u is odd

yangwy 2014-02-05 21:46:07
u must be odd

copeland 2014-02-05 21:46:34
Notice that 81 is about twice all of the possible answer choices. This implies that we may have missed a parity constraint. Indeed, the right side of the equation is always odd, so $u$ has to be odd as well.

copeland 2014-02-05 21:46:43
How many values of $u$ does that leave?

nowunkie 2014-02-05 21:47:20
40!

MSTang 2014-02-05 21:47:20
40

LMA 2014-02-05 21:47:20
40

dli00105 2014-02-05 21:47:20
40

CornSaltButter 2014-02-05 21:47:20
40 values

mathwizard888 2014-02-05 21:47:20
40 (B)

Modest_Ked 2014-02-05 21:47:20
40

brainiac1 2014-02-05 21:47:20
40 values

bobthesmartypants 2014-02-05 21:47:20
40

CornSaltButter 2014-02-05 21:47:20
B!

blueferret 2014-02-05 21:47:20
40

AkshajK 2014-02-05 21:47:20
40

dli00105 2014-02-05 21:47:20
(B)40

WalkerTesla 2014-02-05 21:47:20
40

mathtastic 2014-02-05 21:47:20
40!!

ingridzhang97 2014-02-05 21:47:20
40

joshualee2000 2014-02-05 21:47:20
40

yangwy 2014-02-05 21:47:20
40

simon1221 2014-02-05 21:47:20
40

Acstar00 2014-02-05 21:47:20
40

distortedwalrus 2014-02-05 21:47:20
40

tfmtoto 2014-02-05 21:47:20
40

copeland 2014-02-05 21:47:22
That leaves \[-195,-185,-175,\ldots,175,185,195.\] There are $\boxed{40}$ numbers on this list. The answer must be (B).

copeland 2014-02-05 21:47:26
As upastanding, conscientious mathematicians, we really should check that we can get all of these. Notice for sure that $u=\pm5$ gives solutions since these are our points $A$ and $B.$

copeland 2014-02-05 21:47:34
As rushed and sleepy mathematicians, we probably ought to assume that we are right. Here's a nice compelling picture anyway:

copeland 2014-02-05 21:47:38

copeland 2014-02-05 21:47:49
Notice that we hit the lattice points $A$ and $B$ and these correspond to $u=\pm5.$ We miss the $u=0$ point because the vertex has half-integer $v=\dfrac{-2}5$. Now the rest of the claim is that the solution set to some Diophantine equation corresponds to some arithmetic progression with these two points as consecutive terms. That feels very reasonable and natural to me.

copeland 2014-02-05 21:48:02
In the interest of time, I will leave it to you to figure out what $x$ and $y$ are in terms of $u$ and $v$ (the answer is pretty neat) and then prove that $u=10n+5$ forces $x$ and $y$ to be integers.

Tuxianeer 2014-02-05 21:48:39
upastanding

copeland 2014-02-05 21:48:42
Indeed. We have upas on our farm as well.

copeland 2014-02-05 21:48:52
They help with any legs problems that have sixes in them.

copeland 2014-02-05 21:49:19
Alright, speaking of sleepy. Is anybody else exhausted?

nowunkie 2014-02-05 21:49:52
yes

Modest_Ked 2014-02-05 21:49:52
yes I have school tom

dli00105 2014-02-05 21:49:52
me!!

1915933 2014-02-05 21:49:52
Yes!

bob12345678 2014-02-05 21:49:52
me i should be doing my science fair

mishai 2014-02-05 21:49:52
I should be doing my homework right now.

david_sun 2014-02-05 21:49:52
no, math makes me energized

copeland 2014-02-05 21:50:17
Well I'm getting tired and it's coming up on 7, so I think we should call it for the night.

copeland 2014-02-05 21:50:23
Any questions before we go?

copeland 2014-02-05 21:50:44
There WILL be a transcript for the Math Jam on the Math Jams transcripts page.

copeland 2014-02-05 21:51:03
http://www.artofproblemsolving.com/School/mathjams.php?past=1

copeland 2014-02-05 21:51:09
I'll try to put that up in the next few minutes.

Modest_Ked 2014-02-05 21:51:14
How do you type so fast?

copeland 2014-02-05 21:51:18
I have two keyboards.

simon1221 2014-02-05 21:51:40
and 4 arms

copeland 2014-02-05 21:51:42
There are phenomenal discussions about the rest of the problems in our forum:

copeland 2014-02-05 21:51:43
http://www.artofproblemsolving.com/Forum/viewforum.php?f=133

copeland 2014-02-05 21:51:56
That answers a large number of the questions that many of you have.

DaChickenInc 2014-02-05 21:52:05
Should I look in the last 5 problems of every test to get bookkeeping problems?

copeland 2014-02-05 21:52:20
You will seen lots of bookkeeping problems on the AMC.

copeland 2014-02-05 21:52:39
Practicing these is very useful and you can find them every year.

copeland 2014-02-05 21:52:51
The main trick is to write a lot of examples. Small cases. Etc.

copeland 2014-02-05 21:53:08
Be sure to test where things start and where they end so you don't get off-by-one errors.

fz0718 2014-02-05 21:53:27
Should we make a real, tangible book of tricks?

copeland 2014-02-05 21:53:29
Couldn't hurt.

mathtastic 2014-02-05 21:54:36
is mr rusczyk gonna make videos for the last few problems?

copeland 2014-02-05 21:54:38
Yes!

copeland 2014-02-05 21:54:46
You can find videos for all of the problems we covered today on our videos page:
http://www.artofproblemsolving.com/Videos/index.php?type=amc
Richard chose some different tactics in his videos than we used today, so it might be worth checking them out.

ws0414 2014-02-05 21:55:02
why called bookkeeping?

copeland 2014-02-05 21:55:07
Because it's fun to type.

copeland 2014-02-05 21:55:20
You know that bookkeeping is one of 3 words in the English language with 3 double letters?

DrMath 2014-02-05 21:56:02
i clicked AMC 10B but nothing came up

copeland 2014-02-05 21:56:11
You just jumped to that list on the page.

copeland 2014-02-05 21:58:05
Alright, it looks like it's time to wind this down. thanks to everyone for coming to tonight's Math Jam.

copeland 2014-02-05 21:58:16
I'm going to close the room here in a few moments.

copeland 2014-02-05 21:59:03
Please join us again on Thursday, February 20, when we will discuss the AMC 10B/12B contests and on again in March when we will be discussing the AIME I and II contests.

2014 AMC 10/12 A Discussion

Facilitator: Jeremy Copeland