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Home » Mathematics » Middle School Contests (MATHCOUNTS, AMC 8, MOEMS, Ages 10-13)
 
Mathcounts marathon!!!!
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mathking123
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#1
Mathcounts marathon!!!!
lots and lots of problems(hopefully)
i sometimes read the moems forums and i think that the marathons are such a good idea. so i decided to start one.

the marathon rules(according to me Razz ) are

1) solutions, not just answers, should be posted.
2) the solver of the problem should post 2-3 new problems.
3) no new problems should be posted until all current problems have been solved.
4) rule #3 is more important than rule#2
5) keep the problems mathcounts level (state-national preferred).
6) have fun Razz

ok so this is the first problem:1995 national sprint #4

Mr. Mendez awards extra credit points to his students with quiz grades that exceed the class mean. Given that 107 students take the same quiz, what is the largest number of students who can be awarded extra credit?





PostPosted: Mon Mar 17, 2008 5:01 pm  Back to top 
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Yongyi781
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#2
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106

Extreme case: The first person got a zero, and everybody else (106 of them) got a 100. Then, 106 of them were above the mean, and we can't have 107, so 106 is the answer.


Next problems:

Nationals 1991-1992 Sprint #23:
In now many zeros does 50! end when expressed in standard base ten notation?

Nationals 1991-1992 Target #6:
What is the sum of the prime factors of the number represented by
2^{12} - 2^{11} + 2^{10} - 2^9 + . . . + 2^2 - 2^1?
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Last edited by Yongyi781 on Mon Mar 17, 2008 5:08 pm; edited 2 times in total 
PostPosted: Mon Mar 17, 2008 5:04 pm  Back to top 
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mathking123
Riemann Hypothesis
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#3
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this is my second attempt i better get it right now... ok good, the answer to the first problem is 12, because a 0 at the end is a factor of ten, which is a 2 and a five. for every five that we get, there is at least one "2". so we need to count the number of fives. 50/5 gives 10, but every fifth five is a 25, which has two factors of five. so to count those we do 10/5=2 and now theres no fives left. so the anwerr is 10+2=12 fives.


PostPosted: Mon Mar 17, 2008 5:12 pm  Back to top 
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alanchou
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#4
also, as a note, don't edit your posts to add new questions. let the marathon run smoothly

Solve for x: x^3-11x^2+40x-48=0 (this is probably the hardest question I'll actually post)

big hint

Notice how the LHS can be broken down into x^3-12x^2+48x-64+x^2-8x+16 Does anything look familiar?

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PostPosted: Mon Mar 17, 2008 5:16 pm  Back to top 
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Yongyi781
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#5
alanchou wrote:
Solve for x: x^3 - 11x^2 + 40x - 48 = 0 (this is probably the hardest question I'll actually post)


Basic algebraic manipulation, then:

\begin{align*} (x - 4)^3 + (x - 4)^2 &= 0 \\ (x - 4)^3 &= -(x - 4)^2 \\ x - 4 &= -1 \\ x &= 3 \end{align*}

However, dividing both sides by (x - 4)^2 in the second step left out another root, namely 4.

So \boxed{x = 3\ \text{or}\ 4}
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PostPosted: Mon Mar 17, 2008 5:23 pm  Back to top 
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mathking123
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#6
yongyi, i gotta go do my physics hw, ill be back soon to check out how this is goin
srry....

btw, how do you set your location, rating, and signature??

PostPosted: Mon Mar 17, 2008 5:30 pm  Back to top 
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Yongyi781
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#7
mathking123 wrote:

btw, how do you set your location, rating, and signature??


There's a button called Profile near the top of the page in the bluish box. Click it and you will find out how to set your information Wink
_________________
"quoted and followed by itself is a quine" quoted and followed by itself is a quine.
"yields falsehood when preceded by its quotation" yields falsehood when preceded by its quotation.
(Almost) Red MOP 2009

PostPosted: Mon Mar 17, 2008 5:33 pm  Back to top 
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Ihatepie
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#8
This is a cool one that I found in a Nats test that I saw recently.

Not as hard as it looks. 1992 Sprint Nationals.

Given 7a^2 + 3a^2b + 14ab - 2ab^2 = 61 and
7b^2 - 3a^2b + 2ab^2 =2, find the value of | a + b |.

PostPosted: Mon Mar 17, 2008 6:02 pm  Back to top 
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mathking123
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#9
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adding the two equations together you get 7a^2+7b^2+14ab=63
dividing by 7 and rearranging you get a^2+2ab+b^2=9
which implies (a+b)^2=9
so abs(a+b)=3


ok next problems...
1995 state sprint #7
"a positive three digit integer is divided by a positive two digit integer, yielding an integer qutient and zero remainder. what is the smallest possible integer quotient?"

1996 state sprint #2
"shauna did a number trick with zach. she told him to pick an even number, double it, add 48, divide by 4, subtract 7, mutiply by 2, and ubtract his original number. she then told him the result he should have attained. what was it??

PostPosted: Mon Mar 17, 2008 6:15 pm  Back to top 
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alanchou
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#10
mathking123 wrote:
Click to reveal hidden content
adding the two equations together you get 7a^2+7b^2+14ab=63
dividing by 7 and rearranging you get a^2+2ab+b^2=9
which implies (a+b)^2=9
so abs(a+b)=3


ok next problems...
1995 state sprint #7
"a positive three digit integer is divided by a positive two digit integer, yielding an integer qutient and zero remainder. what is the smallest possible integer quotient?"

1996 state sprint #2
"shauna did a number trick with zach. she told him to pick an even number, double it, add 48, divide by 4, subtract 7, mutiply by 2, and ubtract his original number. she then told him the result he should have attained. what was it??


1

Hey, this time it says positive. Smile Anyways, \frac{198}{99}=2, and there's no way to get 1, so \boxed{2}

2

0\rightarrow 0\rightarrow 48\rightarrow 12\rightarrow 5\rightarrow 10\rightarrow \boxed{10}
In fact, it will always be 10, prove it if you wish

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Get a life and do some math.
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PostPosted: Mon Mar 17, 2008 6:18 pm  Back to top 
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mathking123
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#11
two things

1) if i remember correctly in the first question that i put two posts ago, the actual question in the competition did not say positive. so then the dispute was whether or not to accept 999/-0.1=9990. is 0.1 a two digit number? lolz whatever. and,

2) when you solve a problem please post some more problems...

get it, got it, good... Rotfl Rotfl Rotfl

PostPosted: Mon Mar 17, 2008 6:23 pm  Back to top 
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mathking123
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#12
fine ill post new problems... but whoever solve these should post new problems, to keep the marathon running smoothly

1987 nats individual #5

a math teacher wants to curve a set of test grades so that a student who scored 100 receives a score of 100 , but the student who scored 62 will receive a score of 81. the teacher wishs to use a linear function. find a formula for a linear function f which turns an old grade, x, into a new grade, f(x). write our formula in the form f(x) = mx + b (just trying out the latex). express your answer as the product m *(times) b.

PostPosted: Mon Mar 17, 2008 6:40 pm  Back to top 
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jhredsox93
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#13
I must note that that #23 on a national sprint was also in the 2007 Chapter Countdown round Mr. Green

PostPosted: Mon Mar 17, 2008 6:52 pm  Back to top 
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mathking123
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#14
lol thats funny.
maybe he's not sure about it being from nats
but all of my posts i am 99% sure about dates and problem numbers b/c i am taking this straight from my "all time gratest mathcounts problems" book.. and im pretty sure they have the right dates lol.
does anyone want to try the problems that i posted??

i just learned a little about LATEX (how do people make it slanted??)and im loving it!!!

ax^2 + bx + c = 0

PostPosted: Mon Mar 17, 2008 6:59 pm  Back to top 
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Ihatepie
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#15
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So you have two equations
100=100m + b
81=62m + b

19=38m

m=1/2

100=100(1/2) + b
100=50+b
b=50.

y=1/2 x + 50.




national 1999 sprint #4

After 1/5 of the girls left a school dance, the ratio of girls to boys was 2:3. Then, 44 boys left and the ratio of boys to girls was 2:5. How many students remained at the dance?

PostPosted: Mon Mar 17, 2008 7:01 pm  Back to top 
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mathking123
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#16
after 1/5 of the girls left, the ratio was 2: 3, which is the same as 4: 6
if we "undo" the process of the leaving of the girls we get that before they left the ratio was 5: 6 (girls to boys) lets x be the number of boys at the party and y be the total number of people at the party(before the 44 boys left). Then, we get two equations

x/y = (11 - 5)/(5 + 6) = 6/11, and
(x - 44)/y = 2/(2 + 5) = 2/7

(x - 44)/y = x/y - 44/y,
but x/y = 6/11,
substituting, we get..
6/11 - 44/y = 2/7
44/y = 20/77, so
y = 169.4?, no wait, that can't be right..... Rotfl Rotfl Rotfl

edit: oh, I see what i did wrong...
whats the solution to this problem??
i guess ill post another problem after u post the ans.

PostPosted: Mon Mar 17, 2008 7:17 pm  Back to top 
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Yongyi781
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#17
Ihatepie wrote:
This is a cool one that I found in a Nats test that I saw recently.

Not as hard as it looks. 1992 Sprint Nationals.

Given 7a^2 + 3a^2b + 14ab - 2ab^2 = 61 and
7b^2 - 3a^2b + 2ab^2 =2, find the value of | a + b |.


Just add the two equations together, then the 3a^2b and the 2ab^2 terms cancel out:

7a^2 + 14ab + 7b^2 = 63

Dividing both sides by 7, we see that the left side is a square of a binomial.

\begin{align*} a^2 + 2ab + b^2 &= 9 \\ (a + b)^2 &= 9 \\ a + b &= 3 \\ |a + b| &= \boxed{3} \end{align*}
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"yields falsehood when preceded by its quotation" yields falsehood when preceded by its quotation.
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PostPosted: Tue Mar 18, 2008 11:03 am  Back to top 
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Ihatepie
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#18
Ihatepie wrote:
national 1999 sprint #4

After 1/5 of the girls left a school dance, the ratio of girls to boys was 2:3. Then, 44 boys left and the ratio of boys to girls was 2:5. How many students remained at the dance?


Since no one is answering, I'll provide my own solution.

Click to reveal hidden content
So make the amount of girls left at the end x and the amount of boys at the end y. So at the end, y/2=x/5. Multiplying x from each side, you get y=2x/5.

Working backwards, Since it was 2:3 for girls to boys or 3:2 for boys to girls the step before,
(y+44)/3=x/2

so subsitute what you got for why in the first equation.

(2x/5 + 44)/3 = x/2

Cross multiply:

3x=4x/5 + 88

Multiply each side by 5.

15x=4x + 440.

11x=440

x=40.
Inserting that into the earlier equation: y= 2x/5

y=2(40)/5

y=80/5

y=16.

So the amount of kids left at the end were 40 boys + 16 girls or 56 kids.


New problem:

What is the smallest value of x such that
|5x -1| = |3x + 2|? Express your answer as a common fraction.

PostPosted: Tue Mar 18, 2008 11:57 am  Back to top 
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Yongyi781
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#19
Ihatepie wrote:
What is the smallest value of x such that
|5x -1| = |3x + 2|? Express your answer as a common fraction.


Click to reveal hidden content

So we basically split this problem up into four equations:

\begin{align*} 5x - 1 &= 3x + 2 \\ -5x + 1 &= 3x + 2 \\ 5x - 1 &= -3x - 2 \\ -5x + 1 &= -3x - 2 \end{align*}

But the last two equations are equivalent to the first two when you multiply both sides by -1. So we only have to check the first two equations.

Equation 1:
\begin{align*} 5x - 1 &= 3x + 2 \\ 2x &= 3 \\ x &= \frac{3}{2} \end{align*}

Plug in and check:
\begin{align*} |5(\frac{3}{2}) - 1| &= |3(\frac{3}{2}) + 2| \\ \frac{13}{2} &= \frac{13}{2} \end{align*}

Equation 2:
\begin{align*} -5x + 1 &= -3x - 2 \\ 8x &= -1 \\ x &= -\frac{1}{8} \end{align*}

Plug in and check:
\begin{align*} |5(-\frac{1}{8}) - 1| &= |3(-\frac{1}{8}) + 2| \\ \frac{13}{8} &= \frac{13}{8} \end{align*}

Clearly, \boxed{-\frac{1}{8}} is smaller.
_________________
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"yields falsehood when preceded by its quotation" yields falsehood when preceded by its quotation.
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PostPosted: Tue Mar 18, 2008 1:31 pm  Back to top 
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resurrection
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#20
Or just square both sides...

25x^2-10x+1=9x^2+12x+4
16x^2-22x-3=0
(8x+1)(2x-3)=0
x=\frac{-1}{8} is the desired value.

PostPosted: Tue Mar 18, 2008 2:14 pm  Back to top 
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