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mathboy101
Riemann Hypothesis
Riemann Hypothesis


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#4001
Solution

The maximum product would be of the numbers that are closest together, which are 4, 4, 4, and 4. The product is 4^4, which is \boxed{256}.

New Problem

In a bag of three marbles, there are exactly two blue marbles. If Kia randomly chooses two marbles without replacement, the probability of choosing the two blue marbles is one-third. However, before Kia chooses her two marbles, additional
marbles are added to the bag. The probability of picking two blue marbles without replacement is still one-third. What is the least number of marbles that could be in the bag after the additional marbles have been added?

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PostPosted: Wed Sep 30, 2009 2:50 pm  Back to top 
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goldenboy1.618
Poincare Conjecture
Poincare Conjecture

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#4002
Solution
What we're looking for is a positive integer x such that \frac{a}{x}\times\frac{a - 1}{x - 1} = \frac13.
In these equations, x is the new total number of marbles and a is the number of blue marbles.
Cross multiplying yields 3(a)(a - 1) = x(x - 1).
Testing positive integer values of a, with a > 2, and solving the resulting quadratics for x gives the smallest integer solution triple (a, x) as (6, 10).

A check shows that \frac{6}{10}\times\frac{5}{9} = \frac13, as desired, so the smallest new total is \boxed{10}.


Also, re: the previous solution
This maximum value actually occurs when we use the set \{3, 3, 3, 3, 4\}, which yields the product \boxed{324}.


NP
Triangle ABC has A = 30^\circ, and B is a right angle. In how many distinct ways can I slice ABC into 4 congruent triangles?

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PostPosted: Thu Oct 01, 2009 1:58 pm  Back to top 
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goldenboy1.618
Poincare Conjecture
Poincare Conjecture

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#4003
That problem was probably too hard for MC, but the answer is...
4 ways. It is from an old USAMTS problem (you can check it out at http://www.usamts.org).


NP
A bag contains 12 marbles, 4 blue and 8 red. Lenina draws a marble first, then Bernard, and then John, with no replacements. What is the probability that John picks a blue marble?
A Hint
Start by finding Lenina's probability and Bernard's probability.


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PostPosted: Sat Oct 10, 2009 4:43 am  Back to top 
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garfielddisco123
Poincare Conjecture
Poincare Conjecture


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#4004
Solution
P(LENINA AND BERNARD PICK RED AND JOHN PICK BLUE)=28/165
P(ALL BLUE)=1/55=3/165
P(LENINA AND JOHN PICK BLUE AND BERNARD PICK RED)=4/55=12/165
P(LENINA PICK RED AND BERNARD AND JOHN PICK BLUE)=8/165
(28+8+3+12)/165=51/165

51/165


New Problem
What is the smallest three-digit number in which all of its digits factorial add up to be that number?


PostPosted: Sat Oct 10, 2009 6:45 am  Back to top 
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goldenboy1.618
Poincare Conjecture
Poincare Conjecture

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#4005
Solution
We see that the largest 3-digit factorial is 6! = 720, but anything with that as part of its sum automatically contains a 7 or higher. Thus, we must look for the smallest factorial sum with 5! = 120.

We can start with 120 + 1 + 1 = 122, which doesn't work. Trying small three digit numbers with small factorials yields 145 = 1! + 4! + 5! = 1 + 24 + 120, which works. Our answer is \boxed{145}.


NP
Evaluate the sum of the infinite geometric series 6 + 3.6 + 2.16 + 1.296 + .... (From MC)

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PostPosted: Sat Oct 10, 2009 2:58 pm  Back to top 
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gauss1181
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer


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#4006
The first term is 6 and the common ratio is 0.6.

Sum is \frac{6}{1-0.6}=\frac{6}{0.4}=\boxed{15}.

NP:

Evaluate \frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots
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PostPosted: Sun Oct 11, 2009 6:59 am  Back to top 
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Thunder365
Yang-Mills Theory
Yang-Mills Theory


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#4007
Solution:
\frac {1}{2} + \frac {2}{2^{2}} + \frac {3}{2^{3}} + \frac {4}{2^{4}} + \cdots =

\frac {1}{2} + \frac {1}{2^{2}} + \frac {1}{2^{3}} + \frac {1}{2^{4}} + \cdots +

\frac {1}{2^{2}} + \frac {1}{2^{3}} + \frac {1}{2^{4}} + \cdots +

\frac {1}{2^{3}} + \frac {1}{2^{4}} + \cdots +

\cdots
= 1 + \frac {1}{2} + \frac {1}{4} + \dots = 2


NP:
The polynomial p(x) = x^2 + bx + c has exactly one root r. What is \frac {b}{c} in terms of r?
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PostPosted: Sun Oct 11, 2009 7:33 am  Back to top 
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gauss1181
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer


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#4008
This is a perfect square polynomial.

r^2=c

2r=-b

b=-2r

So \frac{b}{c}=\frac{-2r}{r^2}=\boxed{-\frac{2}{r}}

NP:

The Smith family has 4 children. What's the probability that, given the oldest child is a girl, that there's exactly one boy in the family?
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Latest Problems Typed: 1979 ISL number theory problem

PostPosted: Sun Oct 11, 2009 7:43 am  Back to top 
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Thunder365
Yang-Mills Theory
Yang-Mills Theory


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#4009
Solution:
This is probably wrong...
So we have 3 people left, exactly one of whom must be a boy, and out of 2x2x2 remaining ways to choose the 3 kids, only 3C2=3 includes exactly one boy, so its 3/8?
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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
Last edited by Thunder365 on Sun Oct 11, 2009 12:17 pm; edited 1 time in total 
PostPosted: Sun Oct 11, 2009 7:53 am  Back to top 
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cgyao15
Navier-Stokes Equations
Navier-Stokes Equations


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#4010
I got 1/2 but prolly wrong Razz
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PostPosted: Sun Oct 11, 2009 8:53 am  Back to top 
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randomtext
Hodge Conjecture
Hodge Conjecture

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#4011
Solution
You can pair up 1 with every number up to 26 (excluding itself), and pair up the rest of the numbers with the same difference (1-25) However, you can switch the numbers so that the difference is negative, such as (26,1) instead of (1,26) That gives a total of differences from 1 to 25 and -25 to -1 or \boxed{50} differences.

Sorry, not good at explaining things >>


NP
A hiker can climb a mountain uphill at a speed of 6 mph. He can go downhill at a speed of 10 mph. If he spends no time at the summit, what is his average speed in mph?


PostPosted: Sun Oct 11, 2009 9:02 am  Back to top 
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gauss1181
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer


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#4012
Let x be the distance traveled one way up/down the mountain.

Average speed in miles per hour is \frac{\text{distance}}{\text{time}}=\frac{2x}{\frac{x}{6}+\frac{x}{10}}=\frac{2}{\frac{1}{6}+\frac{1}{10}}=\frac{2}{\frac{4}{... miles per hour.

NP:

Two fair 6-sided dice are rolled. What is the probability that their sum is a double-digit number?
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My Olympiad Compilation! (plus some bits of randomness too)
Latest Problems Typed: 1979 ISL number theory problem

PostPosted: Sun Oct 11, 2009 3:51 pm  Back to top 
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Thunder365
Yang-Mills Theory
Yang-Mills Theory


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#4013
Solution:
We have the possibilities:
2,3,4,5,6,7
3,4,5,6,7,8
4,5,6,7,8,9
5,6,7,8,9,10
6,7,8,9,10,11
7,8,9,10,11,12
so thats a total of 6/36=1/6
@ random text: your solution was wrong. Experiment with some small numbers and you find that the number of POSITIVE (i didnt specify that, sorry) differences will be the factors of n/2.

NP:
Thunder365, athunder, cliu, and Alex Song are going to drive together to Mathcounts. The car they are using has four standard seats: one driver’s seat, one front passenger seat, one left rear seat, and one right rear seat. If Thunder365 and Alex Song are the only ones who can drive, then what is the number of possible seating arrangements?
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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: Sun Oct 11, 2009 3:59 pm  Back to top 
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cgyao15
Navier-Stokes Equations
Navier-Stokes Equations


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#4014
\ 2*3*2?
If wrong correct mua

NP
Shelly find 5 beads. She wants to make a necklace, how many ways can she arrnage them?
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PostPosted: Sun Oct 11, 2009 4:03 pm  Back to top 
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Thunder365
Yang-Mills Theory
Yang-Mills Theory


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#4015
yep its 12.

Solution:
If rotations only matter than its 4!=24. If rotations and flips matter then its 4!/2=12

NP:How many distinct four-digit numbers can be formed by rearranging the four digits in 2009?
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I'm not dumb. I just have a command of thoroughly useless information. Mr. Green
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: Sun Oct 11, 2009 4:08 pm  Back to top 
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cgyao15
Navier-Stokes Equations
Navier-Stokes Equations


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#4016
4!/2=12? i think.

NP
Bob has cut his pizza into 13 slices. 12 are equal but one isn't. What fraction of that pizza is the odd one out?
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PostPosted: Sun Oct 11, 2009 4:10 pm  Back to top 
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$LaTeX$
Riemann Hypothesis
Riemann Hypothesis


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Posts: 471
Location: Why Are You Looking At This STALKER??? Say NO to Stalkers!
ChinaUnited States

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#4017
cgyao15 wrote:


NP
Bob has cut his pizza into 13 slices. 12 are equal but one isn't. What fraction of that pizza is the odd one out?


I don't think there is enough information to solve.
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PostPosted: Sun Oct 11, 2009 4:11 pm  Back to top 
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cgyao15
Navier-Stokes Equations
Navier-Stokes Equations


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#4018
eh solve this then:
I can paint a house in 4 hrs.
Bob can do 3/4 of the same job in 10 hrs.
How long will it take to paint the WHOLE house working together?
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PostPosted: Sun Oct 11, 2009 4:12 pm  Back to top 
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AwesomeToad
Yang-Mills Theory
Yang-Mills Theory


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#4019
Solution
Bob can paint the house in \frac {40}{3} hours and I can paint the house in 4 hours, so I can do \frac {1}{4} of the job per hour and Bob can do \frac {3}{40} per hour. So together, we do \frac {1}{4} + \frac {3}{40} = \frac {13}{40} per hour and it takes \frac {40}{13} hours.


New Problem
Find \sqrt {50\cdot51\cdot52\cdot53 + 1} without a calculator

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PostPosted: Mon Oct 12, 2009 4:02 am  Back to top 
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Math Champion
Yang-Mills Theory
Yang-Mills Theory


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#4020
Solution
We find a pattern:
\sqrt{1\cdot2\cdot3\cdot4+1}=5.
\sqrt{2\cdot3\cdot4\cdot5+1}=11.
\sqrt{3\cdot4\cdot5\cdot6+1}=19.
\sqrt{4\cdot5\cdot6\cdot7+1}=29.

We notice that the difference is 6,8,10,...

Therefore, \sqrt{50\cdot51\cdot52\cdot53+1}=2651.

PS: This problem was from MATHCOUNTS, and so shouldn't be in this marathon.


NP
NP: What is the probability of rolling a sum of 12 when rolling 3 fair 6-sided dice?

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PostPosted: Tue Oct 13, 2009 2:48 pm  Back to top 
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