MATHCOUNTS MATH JAM GAME


Greetings and welcome to today's MATHCOUNTS Math Jam Game. Today we will be examining problems similar to those that might appear on the popular MATHCOUNTS contests.

MCrawford (19:24:41)
Before we get started I would like to take a moment to explain our virtual classroom to those who have not previously participated in a Math Jam or one of our online classes.

MCrawford (19:24:48)
The classroom is moderated meaning that students can type into the classroom, but only the moderators can choose a comment to drop into the classroom. 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.

MCrawford (19:24:57)
Also, only moderators can enter into private chats with other people in the classroom. You can copy the contents of a private chat by clicking on the notebook icon on the private chat window and copying and pasting the contents into a file on your own computer.

MCrawford (19:25:07)
Now, before we get started I would like to explain the game we are going to play today. I will post problems and as soon as I receive three correct responses I will post those responses. Each respondent will score 1 point for a correct answer. Only a student's first submitted answer will count.

MCrawford (19:25:20)
Additionally, the three correct respondents will be allowed to explain the solution. An additional 1 point will be awarded to best solution. If all solutions are equal, the extra point will be awarded to the first of the three students who answered the question correctly. After the correct answers are posted the three respondents will have at most one minute to prepare their solutions. I will let you know when there are ten seconds left.

MCrawford (19:25:30)
One more thing. The problems are all written in TeX and some students report slower loading of the TeX than others. Unfortunately, this is something that I cannot help. Try to enjoy the game and the problems and learn something even if it's not as easy to be one of the first three with the correct answer.

MCrawford (19:25:39)
Now, let's get started!

MCrawford (19:25:47)


SplashD (19:26:20)
-4

tony2smart (19:26:20)
-4

daine (19:26:22)
-4

MCrawford (19:26:33)
Now, for the extra point, explain your solutions.

tony2smart (19:27:09)
this is because 2x+9 will need to be and integer and also an reciprocal so 2x+9=1/(x+5)=1 so x=-4

SplashD (19:27:25)
(2x+9)(x+5)=1 2x^2+19x+44=0 (2x+11)(x+4)=0 since x is integer, x=-4

MCrawford (19:27:57)
The extra point to SplashD.

MCrawford (19:28:06)
Next problem:

MCrawford (19:28:10)
Oops, wait.

MCrawford (19:28:19)
I've prepared my own solutions for many of these.

MCrawford (19:28:23)


MCrawford (19:28:29)


tony2smart (19:28:41)
but isn't my way faster?

tony2smart (19:28:47)
i'm just curious

MCrawford (19:29:10)
I don't even see a complete solution in your work.

shan_ying80@yahoo.com (19:29:15)
i dont get how you got your way

tony2smart (19:29:17)
oooo ok

MCrawford (19:29:26)
Next problem:

MCrawford (19:29:30)


Klebian (19:29:43)
-4

aznness (19:29:43)
-4

mathfreak0388 (19:29:46)
-4

MCrawford (19:30:04)
Lots correct, but these three were first.

MCrawford (19:30:12)
For the extra point, explain how you did this one so fast.

Klebian (19:30:42)
first we see that 5/6 / 2/3 is 5/4. 3/4 + 5/4 is 2, 3/4 - 5/4 is -1/2. 2 divided by -1/2 comes out to be -4.

mathfreak0388 (19:30:46)
5/6 divided by 2/3 is 5/4. 3/4+5/4=2. The denominator is equal to -1/2. 2 divided by -1/2 is /4

aznness (19:30:47)
2/3 = 4/6, and 5/6 divided by 4/6 = 5/4. it becomse 3/4+ 5/4 / 3/4 - 5/4 = 3+5 / 3-5 = 8/-2 = -4.

MCrawford (19:30:59)
The extra point to Klebian.

MCrawford (19:31:06)
Of course, every one of you can do all the computations in this problem. Let's talk about doing them in a way such that you can calculate the answer quickly and have less worries about mistakes.

MCrawford (19:31:15)
First, the fractions in the numerator and denominator are the same. This makes like easier.

MCrawford (19:31:19)


MCrawford (19:31:34)


aznness (19:31:34)
I will vote in favor of the infamous "twitch and buzz in" method

MCrawford (19:31:44)


MCrawford (19:31:53)
Seeing relationships between numbers in simple problems can help us see related structures between numbers when we get to harder problems.

MCrawford (19:32:06)
Next problem:

MCrawford (19:32:08)


kyyuanmathcount (19:32:16)
13

SplashD (19:32:16)
13

daine (19:32:17)
13

MCrawford (19:32:27)
For the extra point, explain your solutions.

SplashD (19:33:10)
first, you can have 4 of each color so that makes 12. then 1 more card will make you definately have 5 of 1 color so it's 12+1=13

daine (19:33:12)
the most you can draw and not have 5 of the same color is 4 of each, 12. add 1 to get 13

kyyuanmathcount (19:33:24)
According to the pigeonhole principle, to be guarenteed k+1 objects, you must have draw nk+1 where n is the number of holes. in this case, k = 4 and n = 3, 3*4 + 1 = 13

MCrawford (19:33:35)
The extra point to kyyuan.

MCrawford (19:33:39)
We solve this problem by drawing the [i]maximum[/i] number of cards [i]without[/i] drawing 5 of the same color. When we have 4 of each card, we have 12 cards. Drawing 1 more means we have 13 total, and 5 of one type.

MCrawford (19:33:47)
This problem is an example of the pigeonhole principle. If you don't know it, you are encouraged to ask about it in the forums.

MCrawford (19:33:55)
Next:

MCrawford (19:33:58)


aznness (19:34:04)
11/21

kyyuanmathcount (19:34:06)
11/21

10SNE1 (19:34:08)
11/21

MCrawford (19:34:21)
For the extra point, explain how you worked that one so fast.

10SNE1 (19:34:29)
the p(red)=7/21

10SNE1 (19:35:20)
the p(red)=7/21, the p(green)=3/21, p(blue)=1-p(red+green)=1-10/21=11/21

kyyuanmathcount (19:34:46)
You must either draw red green or blue. The red is 7/21 while the green is 3/21.
(3+7)/21 = 10/21, so the remaining 11/21 must be the probability of drawing blue.

aznness (19:35:14)
I twitched. 1/7 = 3/21. 1/3 =7/21. 3+7 = 10. All i saw were 3 colors and 2 probabilities, so i assumed the other color was the one left. I highly disrecommend you do this in countdown, as you may misread and give your opponent up to 40 seconds to think it over.

MCrawford (19:35:35)
The extra point to axnness for pure twitchiness.

MCrawford (19:35:51)
Next problem:

MCrawford (19:35:52)


aznness (19:35:55)
1/6

kyyuanmathcount (19:35:56)
1/6

Klebian (19:35:57)
1/6

MCrawford (19:36:14)
For the extra point, explain your solutions.

aznness (19:36:44)
x/7 uses the 6 digits 142857 in a block. any of the digits has a 1/6 probability. Just hope you don't miss the 6 key. May the Twitch be with you.

Klebian (19:36:49)
Every representation of x/7 repeats every 6 digits. Since 1/7 is .142857..., and each x/7 is just a different order, the P of getting a 4 is 1 out of 6

kyyuanmathcount (19:37:04)
The decimal representation of x/7 will have a repeating strand of 6 digits (142857) because you can at most have 6 different non-divisible modulo residues for a repeating decimal. thus, a 4 is one of the six digits, 1/6

MCrawford (19:37:24)
The extra point to aznness for quick draw.

MCrawford (19:37:29)
This problem isn't particularly difficult for students who have an understanding of repeating decimals. However, there are a few things we can note that make the problem even easier.

MCrawford (19:37:36)
First, all reduced fractions with denominator 7 have the same digits (in the same cycling order) in their repeating decimal. We know this because when we do the long division to find the repeating decimal for 1/7, all 6 possible remainders appear once. If they didn't, the decimal would have repeated before 6 digits. This means that the fractions 1/7, 2/7, 3/7, 4/7, 5/7, and 6/7 all have the same repeating digits in the same order except with different starting points.

MCrawford (19:37:45)


MCrawford (19:37:53)
Since 4 is one of the six digits in any of these repeating decimals, the answer is 1/6.

Zmastr (19:38:07)
would it change based on the 0?

MCrawford (19:38:21)
I'm not sure what you're asking.

MCrawford (19:38:43)
But questions about the solutions are best left for the forums so that we can play more game here.

Zmastr (19:38:39)
Would the 0 influence the probability, since it's one of the numbers in the decimal

MCrawford (19:38:54)
It's not part of the repeating digit block.

MCrawford (19:39:05)
Next problem:

MCrawford (19:39:06)


SplashD (19:39:19)
9

daine (19:39:22)
9

xxm0rg07hxx (19:39:40)
9

MCrawford (19:39:51)
For the extra point, explain your solutions.

xxm0rg07hxx (19:40:23)
There boundaries for obtaining a negative solution are -5 and 3 so 9 integers.

SplashD (19:40:59)
x^2+2x-19=(x+1)^2-20 for it to be negative, (x+1)^2 has to be less than 20 so (x+1) can be any integer from -4 to 4 inclusive and that's 9 integers

MCrawford (19:41:47)
The extra point to SplashD.

MCrawford (19:41:55)
Really quick MATHCOUNTS students can get this problem as quickly as they can read it. Let's take a look at the reasoning behind their intuition.

MCrawford (19:42:02)


MCrawford (19:42:13)


MCrawford (19:42:29)
The endgame is quick and easy. We want squares less than 20, and their square roots, including negatives. The integers between -4 and 4 inclusive can be values of y, so 9 there are 9 values of y corresponding to 9 values of x.

tony2smart (19:42:28)
the solution are -5 to 3 though...

MCrawford (19:42:47)
Sure, when we shift back for the variable x.

MCrawford (19:43:07)
Next problem:

MCrawford (19:43:08)


kyyuanmathcount (19:43:27)
18 rt(3)

Someone (19:43:33)
18 rt.3

SplashD (19:43:39)
18 cube root(3)

MCrawford (19:43:49)
For the extra point, explain your solutions.

aznness (19:43:59)
splshd is wrong...

MCrawford (19:44:19)
Oops, true. aznness was next.

MCrawford (19:44:43)
Sorry SplashD.

MCrawford (19:45:03)
I'm waiting for solutions from aznness and kyyuan.

Someone (19:44:14)
They form an equilateral triangle with side length 6 rt.2. using formula rt.3 x^2/4, we get 18 rt.3

kyyuanmathcount (19:45:02)
Because A B and C are connected by ALL face diagonals, the riangle is equilateral. Since the volume is 216, the side is cubert(216)=6.
[6rt(2)]^2 rt(3)/4 = 18 rt(3)

aznness (19:45:27)
sorry splash. The only way for the three vertices to not be connected by edges is for them to be connected by face diagonals. Since all three are connected by face diagonals, the side is 6sqrt2. we can plug this into equilateral triangle formula and get 18sqrt3

MCrawford (19:45:43)
The extra point for kyyuan.

MCrawford (19:46:05)
Could those of you with moving avatars please remove them. They are distracting.

MCrawford (19:46:07)
Thanks.

MCrawford (19:46:13)
The only way to select three corners of a cube such that no two share an edge is if each is opposite a face diagonal from the other.

MCrawford (19:46:18)


MCrawford (19:46:42)
Next problem:

MCrawford (19:46:43)


aznness (19:46:50)
7/20

Someone (19:47:05)
7/20

SplashD (19:47:10)
7/20

MCrawford (19:47:27)
Props to aznness who got this one in about 2 seconds.

MCrawford (19:47:32)
For the extra point, explain your solutions.

aznness (19:48:20)
1/x^2 + 1/y^2 = (x^2 + y^2)/(x^2y^2). X^2 + y^2 = (x-y)^2 + 2xy. x^2y^2.....

the numerator is thus 140, and the denominator is 400. 140/200=7/20

Someone (19:48:23)
the fraction thingy over a common denom becomes y^2+x^2/x^2y^2, the bottom is the same as (xy)^2 or 400. and the top is the same as (x-y)^2+2xy, or 140. taking 140/400 and simplifying we get 7/20

SplashD (19:48:39)
(1/x^2)+(1/y^2)=(x^2+y^2)/(x^2y^2)=((x-y)^2+2xy)/(9xy0^2)=140/400=7/20

MCrawford (19:48:54)
The extra point to aznness for being quick on the draw.

MCrawford (19:49:04)


MCrawford (19:49:18)
Many algebraic manipulation problems involve a product of two variables along with their sum or difference because such an enormous number of algebraic expressions can be built from them. Some of you may have already studies more general techniques for solving "symmetric polynomial" problems.

MCrawford (19:49:40)
Often, if you just keep looking for new ways to simplify or express these kinds of problems, you eventually find a way to plug the numbers in.

MCrawford (19:49:51)
Next problem:

MCrawford (19:49:52)


kyyuanmathcount (19:50:04)
1/9

Klebian (19:50:04)
1/9

SplashD (19:50:05)
1/9

MCrawford (19:50:50)
For the extra point, explain your solutions.

MCrawford (19:51:16)
Testing.

MCrawford (19:51:27)
Is anybody reading this?

DPatrick (19:51:54)
We're having a slight technical problem -- please bear with us.

MCrawford (19:52:55)
Who were the three who had solutions that I posted?

Zmastr (19:53:10)
kyyuan, klebian, and splashd

kyyuanmathcount (19:50:56)
Lets call Naoki's probability x. Then dave has probability 2x, and Richard has probability 6x. x+2x+6x=9x. x/9x = 1/9
(AoPS people, yay)

Klebian (19:51:26)


SplashD (19:51:46)
let the probability that Naoki will win be x, Dave, 2x and Richard 6x. So the probability Naoki will is x/9x=1/9

MCrawford (19:53:48)
Who was first?

kyyuanmathcount (19:53:53)
me

MCrawford (19:54:04)
The extra point to kyyuan.

MCrawford (19:54:22)
At the halfway point, we have a close game.

MCrawford (19:54:35)
aznness and kyyuanmathcount have 8 points each.

MCrawford (19:54:39)
SplashD has 7.

MCrawford (19:54:50)
Klebian has 4 points and daine has 3.

Klebian (19:54:37)
the twitch master in the lead

MCrawford (19:55:04)
Most probability problems we see in MATHCOUNTS are solved using counting or geometric techniques. This problem is different. We solve it using algebra.

MCrawford (19:55:11)
Let R be the probability that Richard wins, D be the probability that Dave wins, and N be the probability that Naoki wins. We have three equations:

R + D + N = 1
R = 3D
D = 2N

MCrawford (19:55:21)


aznness (19:55:10)
twitch master tied with guy with same initials. GRR IM KYY too.

MCrawford (19:55:34)
Here is an article I recommend to MATHCOUNTS students who can solve most MATHCOUNTS probability problems, but want to reach into hard problems. The focus of the article is on distinguishing between different types of probability problems:

MCrawford (19:55:36)
http://www.artofproblemsolving.com/Resources/probarticle.pdf

kyyuanmathcount (19:55:37)
Im actually AY, KYY is my dad

MCrawford (19:55:55)
Next problem:

MCrawford (19:55:58)


kyyuanmathcount (19:56:01)
15

Zmastr (19:56:15)
15

math1001 (19:56:15)
15

MCrawford (19:56:27)
For the extra point, explain how you three did that so fast.

kyyuanmathcount (19:56:39)
3*65 = 195. He would average 60 mph if the trip took 180 minutes, thus, 195-180=15

math1001 (19:57:04)
he takes 3*65=195 normally, and 180 at 60 mph.195-180=15

Zmastr (19:57:38)
On a regular trip he goes at 65 mph for 3 hours. Therefore, the trip is 195 miles long. If he averages 60 mph, then he only spends 180 on the road. There is 15 minutes left over.

MCrawford (19:58:03)
The extra point to kyyuan for speed.

MCrawford (19:58:10)


MCrawford (19:58:20)
Make sure you see how that works so that you can apply it readily to problems.

MCrawford (19:58:36)


kyyuanmathcount (19:58:49)
60

aznness (19:58:52)
60

mathfreak0388 (19:58:55)
60

MCrawford (19:59:04)
For the extra point, explain your solutions.

aznness (19:59:55)
The other endpoint is (6,4), as can be calculated with reverse-midpoint formula, and as the sides are parallel to the axes, one side is 6 - -4=10, the other is 4--2=6. thus 6*10=60

kyyuanmathcount (20:00:02)
The endpoints of the diagonal and the intersection of the diagonal's coordinate difference for both x and y are half the sidelength.
For x, 1- (-4) = 5, so one side is 10. The other side is 2 - (-1) = 3, 3*2 = 6, thus, 6*10=60

mathfreak0388 (20:00:02)
The distance from the x-coordinates are half the length of the rectangle, so the lenght is 10. The same for the y-coordinates, which adds up to 6. 6x10=60

MCrawford (20:00:34)
The extra point to kyyuan.

MCrawford (20:00:43)
The key to solving this problem quickly is recognizing that the rectangle with coordinates (-4, 2), (1, 2), (1, -1), and (-4, -1) has 1/4 the area of the rectangle we want. The area of this rectangle is (1 - (-4))(2 - (-1)) = 5 x 3 = 15. Our answer is 4(15) = 60.

MCrawford (20:01:09)
Next problem:

MCrawford (20:01:11)


MCrawford (20:02:23)
10 seconds...

aznness (20:01:53)
15

kyyuanmathcount (20:02:07)
15

MCrawford (20:02:47)
For the extra point, explain your solutions.

aznness (20:03:48)
use casework.
Nothing- 1.
Only 2s- 6 2,4,8,16,32,64
Only 5s- 5,25
2 and 5s- 10,50
4 and 5s- 20, 100
8 and 5- 40
16 and 5- 80.

and that's all the cases, adding up to 15

kyyuanmathcount (20:04:02)
Integers that have terminating representations have only factors of two or five. The ones with two are 2^1, 2^2 ....2^6
with both two and five, we have 2*5, 2*25, 4*5, 4*25 etc. for a total of 15 cases

MCrawford (20:04:17)
The extra point to aznness for quickness.

MCrawford (20:04:31)
Now, let's discuss a solution that organizes the casework to make it a little easier.

MCrawford (20:04:37)
A rational number has a terminating decimal if and only if the prime factorization of the denominator includes no primes other than 2 or 5. If you don't already know why this is the case, think about the fact that in the decimal system, we can multiply any terminating decimal by a large enough power of 10 to get an integer.

MCrawford (20:04:48)
Now, our goal is to count the positive integers less than 125 that have no prime divisors besides 2 or 5.

MCrawford (20:04:53)
Casework helps us organize the process.

MCrawford (20:04:59)


MCrawford (20:05:11)


MCrawford (20:05:21)


MCrawford (20:05:32)


MCrawford (20:05:44)
Adding the counts from each case, we get our answer:
7 + 5 + 3 = 15.

MCrawford (20:06:05)
Prime factorizations help us organize information about divisors.

MCrawford (20:06:11)
Next problem:

MCrawford (20:06:12)


aznness (20:06:22)
1013

kyyuanmathcount (20:06:23)
1013

tony2smart (20:06:30)
1013

MCrawford (20:06:40)
For the extra point, explain your solutions.

tony2smart (20:07:03)
2^10 is all the possible commitees - 10 for one person-1 for no one= 1024-10-1=1013

aznness (20:07:12)
You have a total of 2^10 committees (each person choosing in or out) 0 is invalid, has 1. 1 is invalid, has 10. 1024-1-10=1013

kyyuanmathcount (20:07:19)
There are 2^10 commitees of any size (Each person can be in or out). The ones with one person can be selected in 10 ways, and there is one commitee with zero people. 1024-11=1013

MCrawford (20:07:37)


MCrawford (20:07:46)
The extra point to aznness for being quick on the draw.

MCrawford (20:07:57)
Close game. kyyuan leads by 1.

MCrawford (20:08:08)
Next problem:

MCrawford (20:08:09)


mathfreak0388 (20:08:25)
1/2

kyyuanmathcount (20:08:27)
1/2

SplashD (20:08:30)
1/2

MCrawford (20:08:50)
For the extra point, explain your solutions.

mathfreak0388 (20:09:02)
1/5x(x)=1/10. solving, x=1/2

kyyuanmathcount (20:09:07)
call the fraction x. then, 2x^2/5 = 1/10. Solving this equation yields x^2=1/4 so x=1/2.

SplashD (20:09:23)
(2/5)x^2=(1/10) x^2=1/4 since x is positive, x=(1/2)

MCrawford (20:09:39)
The extra point to mathfreak.

MCrawford (20:09:56)
Next problem:

MCrawford (20:09:57)


aznness (20:10:06)
-6

Klebian (20:10:09)
-6

daine (20:10:10)
-6

MCrawford (20:10:30)
For the extra point, explain your solutions.

Klebian (20:10:56)
first we see that |x+3| = 7. so x+3 can equal 7 or -7. When we take the first case, x is 4. In the second case, x is -10. 4 + -10 is -6

aznness (20:11:04)
Ix+3I = 7. X is either 4 or -10. 4-10=-6

daine (20:11:20)
the |x+3|=7 so x+3 = 7 or -7 x=-10,4

MCrawford (20:11:32)
The extra point to aznness.

kyyuanmathcount (20:11:28)
Well, it doesn't matter what's on the right side.

MCrawford (20:11:43)
Exactly!

MCrawford (20:11:49)
The symmetry in the graph is around -3.

MCrawford (20:12:00)
Well, we do need for the equation to have more than 1 solution.

MCrawford (20:12:14)
There could be 0 or 1 solution, depending on the numbers.

MCrawford (20:12:37)
But since we have an absolute value equal to a positive number, they sum around x = -3.

MCrawford (20:12:55)
Next problem:

MCrawford (20:12:57)


aznness (20:13:04)
540

b-flat (20:13:04)
540

kyyuanmathcount (20:13:07)
540

MCrawford (20:13:23)
For the extra point, explain how you did that so fast.

kyyuanmathcount (20:13:41)
x = x(x-3)/2, so x=5. (5-2)(180)=540

aznness (20:13:53)
x= (x-3)x/2. Or, if you know your regular polygons, x=5. The interior sum is equal to (x-2)180, which is 3*180=540

b-flat (20:14:17)
Write an equation n(n-3)/2=n, n=5, there are (5-2)180=540 degrees. I had memorized now many diagonals common polygons have.

MCrawford (20:14:42)
The extra point to aznness who moves into the lead with 17 points!

MCrawford (20:14:54)
Next problem:

MCrawford (20:14:56)


mathfreak0388 (20:15:06)
5

10SNE1 (20:15:06)
5

Klebian (20:15:07)
5

MCrawford (20:15:16)
For the extra point, explain your solutions.

mathfreak0388 (20:15:29)
Adding equations u get 2b^2=50 b=5

Klebian (20:15:53)
adding up both equations, we get 2a^2 = 110. that means a^2 = 55. Subtracting 55 from the 80 in the first equation, we get b^2=25. That means b=5, since it is postive

10SNE1 (20:15:55)
subtract the equations to get 2b^2=50, b^2=25, so b=5

MCrawford (20:16:23)
Though he said adding instead of subtracting, the extra point to mathfreak0388 for going straight to the solution.

MCrawford (20:16:25)


MCrawford (20:16:31)
Since b > 0, the answer is 5.

MCrawford (20:16:47)
Next problem:

MCrawford (20:16:49)


biffanddoc (20:17:07)
7/10

aznness (20:17:09)
7/10

kyyuanmathcount (20:17:10)
7/10

MCrawford (20:17:20)
For the extra point, explain your solutions.

kyyuanmathcount (20:18:18)
The probability that the median is equal to three is 4/10 (the other numbers can be 24,15,14,25 and there are 5C3 = 10 ways to choose).
1-4/10 = 6/10. Since the probability of each side is symmetrical, 3/10+ 4/10 = 7/10

aznness (20:18:20)
In order to get 3, you can pick one number of 2 less than and one number of 2 more than. This results in 4. To get 4, you can must have 5 and one of 3 less than, resulting in 3. 4+3 / 4+3+3 (2=4) gives you 7/10

MCrawford (20:19:16)
The extra point to kyyuan for the solution I understand.

biffanddoc (20:19:23)
Complementary counting is the best way here. You have to find sets of 3 numbers with the average being below 3, or the sum 8 or below. The ways are (1,2,3),(1,2,4),(1,2,5),(1,3,4). but that makes the answer (10-4)/10 so isnt it 6/10=3/5?

MCrawford (20:19:39)
nvm, extra point to biffandoc, he was just a little late posting.

MCrawford (20:19:53)
Probably the easiest way to approach this problem is using "complementary counting" or "complementary probability".

MCrawford (20:20:00)
The only way that the median (middle) of the three numbers is less than 3 is if the two smaller numbers are 1 and 2. There are 3 possible values for the largest number, so there are 3 possible ways for the median to be less than 3.

Klebian (20:19:59)
wait... but his solution is wrong.

MCrawford (20:20:18)
Oh, right. Extra point to kyyuan again.

MCrawford (20:20:27)


MCrawford (20:20:31)


biffanddoc (20:20:21)
which one of my sets of numbers doesnt work?

MCrawford (20:20:52)
{1, 3, 4}

MCrawford (20:21:03)
Next problem:

MCrawford (20:21:04)


MCrawford (20:22:28)
Hold on. A spammer posted dozens of messages and I'm trying to find answers.

kyyuanmathcount (20:21:26)
150

SplashD (20:21:27)
150

MCrawford (20:23:11)
For the extra point, explain your solutions.

kyyuanmathcount (20:23:06)
|x|+|y| <= 10 is a square with side 10rt(2). Each of the lines x=-5 and x=5 lop of 1/4 of the area of their respective halves of the x axis, thus, (10rt(2))^2 * 3/4 = 150.

SplashD (20:23:42)
the first equation gives us a square with side length 10 rt 2. the second equation cuts off two isosoles triangles with legs 5 rt 2. so the area is 200-50=150

MCrawford (20:24:09)
The extra point to kyyuan, who moves back into the lead with 20 points.

MCrawford (20:24:23)
Last problem:

MCrawford (20:24:24)


aznness (20:24:33)
1/4

kyyuanmathcount (20:24:40)
1/4

10SNE1 (20:24:42)
1/4

MCrawford (20:24:52)
For the extra point, explain your solutions.

aznness (20:25:26)
F(-2) 1/ (1-3/2) = -2. So, you just need -2^-2. Divide by -2 3 times and get 1/4

kyyuanmathcount (20:25:48)
Plugging -2 in the function and working from the bottom, we see it ends up equaling -2.
Since functions only have one output for each input, sticking it into the function again only makes it -2 again.
(-2)^(-2)=1/4

10SNE1 (20:26:02)
f(-2)=-2, so f(-2)=-2. -2^-2=1/(-2^2)=1/4

MCrawford (20:26:10)
The extra point to kyyuan.

MCrawford (20:26:30)
Now, let's look at the top point scorers today:

MCrawford (20:26:38)
10SNE1 had 3 points.

MCrawford (20:26:44)
daine had 4 points.

MCrawford (20:26:53)
Klebian had 6 points.

MCrawford (20:27:00)
mathfreak0388 also had 6 points.

MCrawford (20:27:08)
SplashD finishes 3rd with 9 points.

MCrawford (20:27:19)
In second with 19 points is aznness.

MCrawford (20:27:40)
Our MATHCOUNTS Math Jam Game champion is kyyuanmathcount with 22 points.

aznness (20:26:27)
=(

Klebian (20:27:29)
nice LARGE gap there.

MCrawford (20:27:53)
This problem relates to a function that is an example of a special kind of function.

MCrawford (20:27:57)


MCrawford (20:28:01)
As it turns out f(x) = x once we wade through all the algebra. Since repeated applications of g to a variable x results in x, we call g a "cyclic function".

MCrawford (20:28:07)


Zmastr (20:27:46)
There's no way they can top my amazing score of 1 point!

kyyuanmathcount (20:28:03)
[img id=em-10] good game everyone!!!

MCrawford (20:28:16)
That's all for today's MATHCOUNTS Math Jam Game. I hope you enjoyed it.

Good luck to those of you competing at nationals next month!

MCrawford (20:28:22)
Note that you can find a transcript of this Math Jam and past Math Jams here:

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php

This Math Jam will be posted later today.

slamchess (20:28:16)
good job every one

Klebian (20:28:27)
who here IS going to nationals?

Zmastr (20:28:30)
who was the spammer you were talking about?

b-flat (20:28:32)
Thank you for hosting the class!

ch1n353ch3s54a1l (20:28:33)
gj everyone

mathfreak0388 (20:28:33)
bye evryone

weiwei (20:28:33)
good job everyone

ch1n353ch3s54a1l (20:28:39)
Thank you

weiwei (20:28:42)
good job anzness

kyyuanmathcount (20:28:43)
~~~THANK YOU!~~~