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Mathcounts marathon!!!!
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pytheagle
Riemann Hypothesis
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#5941
Solution


By definition of probability, we have P=\tfrac{P\text{(success)}}{P\text{(possible)}}. Since we have two choices for every toss namely heads or tails, we have P\text{(possible)}=2^{10}=1024.

Let H be heads, T be tails, and _ represent a toss. Then, the desired outcomes are presented as follows:

1) H H H T _ _ _ _ _ _
2) T H H H T _ _ _ _ _
3) _ T H H H T _ _ _ _
4) _ _ T H H H T _ _ _
5) _ _ _ T H H H T _ _
6) _ _ _ _ T H H H T _
7) _ _ _ _ _ T H H H T

Notice that for all seven of the cases, we need to choose either H or T for the remaining six tosses leading to P\text{(success)}=7\times2^5=224.

Hence, our answer is 224/1024=\boxed{7/32}.



New Problem


A student's scores on five mathematics tests were 83, 67, 79, 92, and 70. The student needs an average score of eighty-one to receive a B grade on the report card. What is the lowest score the student can obtain on the one remaining test before report cards and still receive a B grade? (Source: MATHCOUNTS)


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PostPosted: Fri Oct 09, 2009 7:38 pm  Back to top 
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greenhand
Hodge Conjecture
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#5942
pytheagle wrote:
Solution


By definition of probability, we have P = \tfrac{P\text{(success)}}{P\text{(possible)}}. Since we have two choices for every toss namely heads or tails, we have P\text{(possible)} = 2^{10} = 1024.

Let H be heads, T be tails, and _ represent a toss. Then, the desired outcomes are presented as follows:

1) H H H T _ _ _ _ _ _
2) T H H H T _ _ _ _ _
3) _ T H H H T _ _ _ _
4) _ _ T H H H T _ _ _
5) _ _ _ T H H H T _ _
6) _ _ _ _ T H H H T _
7) _ _ _ _ _ T H H H T

Notice that for all seven of the cases, we need to choose either H or T for the remaining six tosses leading to P\text{(success)} = 7\times2^5 = 224.

Hence, our answer is 224/1024 = \boxed{7/32}.



New Problem


A student's scores on five mathematics tests were 83, 67, 79, 92, and 70. The student needs an average score of eighty-one to receive a B grade on the report card. What is the lowest score the student can obtain on the one remaining test before report cards and still receive a B grade? (Source: MATHCOUNTS)



solution

81*6-70-92-79-67-83=95

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PostPosted: Sat Oct 10, 2009 5:57 am  Back to top 
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garfielddisco123
Poincare Conjecture
Poincare Conjecture


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#5943
Hi do not look, answer ans solution
(83+67+79+92+70+x)/6=81
83+67+79+92+70+x=486
391+x=486
x=95


PostPosted: Sat Oct 10, 2009 6:30 am  Back to top 
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gaussintraining
Riemann Hypothesis
Riemann Hypothesis

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#5944
New Problem

The parallel sides of an isosceles trapezoid have lengths of 19 units and 29 units, and the perimeter is 74 units. What is the area of the trapezoid?


PostPosted: Sat Oct 10, 2009 7:33 am  Back to top 
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AndrewTom
Navier-Stokes Equations
Navier-Stokes Equations

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#5945
Solution


Length of equal sides = (74 - 19 - 29)/2 = 13. Perpendicular height = 12, by Pythagoras.

Area = 12(29+19)/2 = 288.



New problem


Prove that ax^2 + bx + c takes integral values for all values of x if and only if 2a, a+b and c are each integers.



PostPosted: Sat Oct 10, 2009 10:55 am  Back to top 
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pytheagle
Riemann Hypothesis
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#5946
The general term "MATHCOUNTS Level" does not extend to include problems in which proofs are required.

New Problem


How many strings of 3 letters do not include a B?


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PostPosted: Sat Oct 10, 2009 11:55 am  Back to top 
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cgyao15
Navier-Stokes Equations
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#5947
\ 25^{3} go calculate Razz

NP
Find number of ways to spell: cgyao15cyao15 as 1 word.

PostPosted: Sat Oct 10, 2009 12:17 pm  Back to top 
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goldenboy1.618
Poincare Conjecture
Poincare Conjecture

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#5948
Solution
Assuming you meant cgyao15cgyao15, we have 14 letters to be arranged in 14! permutations. Dividing by 2!^7 for repetition gives \frac{14!}{128} = \boxed{681080400}.


NP
If Kelly is 36 years old, then Mark is 29 years old. If Kelly is not 36, then Andrea is not 14. Andrea is indeed 14. Mark is either 29 or 45. How old is Mark? (From a school entrance exam)

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PostPosted: Sat Oct 10, 2009 3:18 pm  Back to top 
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Maybach
Navier-Stokes Equations
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#5949
Solution


Um.....REALLY EASY

This means Kelly is 36 and Mark is 25. (Wow, some school entrance exam Rotfl )


NP

In Texas, 10 students on the 2009 Mathcounts State Competition got 46/46. If only 4 people can make the national competition and one of them was already determined, how many ways can the remaining 3 people be selected from the remaining students?

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PostPosted: Sat Oct 10, 2009 3:39 pm  Back to top 
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the cliu
Poincare Conjecture
Poincare Conjecture


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#5950
solution
Since one of them was already determined, there are 9 students left to distribute to 3 spots on the National Math Counts team. Since order doesn't really matter, our answer is \binom {9}{3} = \boxed {84} ways.


new problem
The probability that cliu will eat Chinese food for dinner is \frac {8}{9}. The probability that cliu will eat Chinese food for lunch is \frac {2}{13}. What is the probability that cliu will not eat Chinese food for lunch or dinner?


PostPosted: Sat Oct 10, 2009 7:05 pm  Back to top 
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Maybach
Navier-Stokes Equations
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#5951
Solution

Complementary counting:
1/9 * 11/13 = 11/117


NP

The probability that Maybach will have Chinese food for lunch is 1/2,500. The probability Maybach will have Chinese food for dinner is 1/360. What's the probability Maybach will have Chinese food for both lunch and dinner?(I always feel sick to my stomach after eating spicy food)

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PostPosted: Sun Oct 11, 2009 5:25 am  Back to top 
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greenhand
Hodge Conjecture
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#5952
solution

P=(1-\frac{8}{9})(1-\frac{2}{13})=\frac{11}{117}


new problem

Find the focus and the directrix of the parabola y=ax^2+bx+c

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PostPosted: Sun Oct 11, 2009 5:32 am  Back to top 
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AndrewTom
Navier-Stokes Equations
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#5953
Mathycounts marathon!!!!

This seems rather difficult for this level.

Solution


Comparing with y=x^{2}, which has its focus at (0, \frac{1}{4}) and directrix y=\frac{-1}{4}, by completing the square to get y=ax^{2} + bx +c in the form (y-y_{0}) = 4(\frac{1}{4a})(x-x_{0})^{2}, I get the focus at (\frac{-b}{2a}, \frac{1}{4a} + \frac{4ac-b}{4a}) and the directrix as y= \frac{-1}{4a} +\frac{4ac-b^{2}}{4a}



New problem


Make 100 with five 1s.



PostPosted: Sun Oct 11, 2009 6:34 am  Back to top 
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Thunder365
Yang-Mills Theory
Yang-Mills Theory


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#5954
lol
Solution
100=111-11


NP
Carla is writing all the numbers from 0 to 99999 inclusive on a piece of paper. How many times does the digit "1" appear?

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PostPosted: Sun Oct 11, 2009 7:02 am  Back to top 
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cgyao15
Navier-Stokes Equations
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#5955
@goldenboy i didn't make a typo, cyao15 is a player too you know Razz
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PostPosted: Sun Oct 11, 2009 3:47 pm  Back to top 
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goldenboy1.618
Poincare Conjecture
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#5956
@cgyao15 Sorry for the confusion.

Solution
She writes 100,000 total numbers. This means she writes 1 in the units place 10,000 times, in the tens place 10,000 times, and so on for each place value, giving her a total number of 5\times10,000 = \boxed{50,000} ones.


NP
What is the degree measure of the smaller angle formed by a clock's hands at 8:20?

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PostPosted: Wed Oct 14, 2009 4:35 am  Back to top 
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AwesomeToad
Yang-Mills Theory
Yang-Mills Theory


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#5957
Solution
The minute hand is 120 degrees past the 12 and the hour hand is 240+\frac{1}{3}30=250 so the smaller angle formed is 250-120=130 degrees

New Problem
Find all real solutions of |x^2-3x|=4

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PostPosted: Wed Oct 14, 2009 8:21 am  Back to top 
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AndrewTom
Navier-Stokes Equations
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#5958
Solution


Sketching the graph, x = - 1 or x = 4.

Alternatively, square both sides and factorise to get (x - 4)(x + 1)(x^2 - 3x + 4) = 0, giving the real roots x = 4, x = - 1.



NP


Solve \mid x - 2 \mid + \mid 1 - x \mid = 0.



PostPosted: Wed Oct 14, 2009 10:19 am  Back to top 
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gauss1181
Birch & Swinnerton Dyer
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#5959
The absolute values must be both 0. But that's not possible.

So there are \boxed{\text{no solutions}} in this equation.

NP:

Find the radius of the incircle of a triangle with side lengths 7, 24, and 25.
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PostPosted: Wed Oct 14, 2009 10:37 am  Back to top 
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Thunder365
Yang-Mills Theory
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#5960
Solution:
A=RS
7-24-25 is a right triangle, so A=84. S=(7+24+25)/2=28. 84/28=3

NP:
Six students are going to sit in a row for Countdown Round in Mathcounts. Thunder wants to sit next to cliu. Athunder does not want to sit next to Alex. The other two people can sit anywhere. How many ways can these six students be seated?
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Goals: AMC8:25| AMC10:138+|AIME:5+| Mandelbrot:Leaderboard| Mathcounts:Make Nats| MMPC part 1:28+, MMPC part 2:16+, top 100=>Michigan ARML B team| USAMTS:Bronze| Make CMO

PostPosted: Wed Oct 14, 2009 1:34 pm  Back to top 
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