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Home » Mathematics » Middle School Contests (MATHCOUNTS, AMC 8, MOEMS, Ages 10-13)
 
MOEMS Marathon
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Math Champion
Yang-Mills Theory
Yang-Mills Theory


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#3981
Solution

The answer is 0, because there will always be at least 1 obtuse triangle out of the 4 that are formed.


NP

A rectangular garden 50 feet long and 10 feet wide is enclosed by a fence. To
make the garden larger, while using the same fence, its shape is changed to a
square. By how many square feet does this enlarge the garden?
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PostPosted: Tue Sep 15, 2009 5:57 pm  Back to top 
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goldenboy1.618
Poincare Conjecture
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#3982
Solution
The original garden has area 50*10 = 500. The new garden has area 30*30 = 900. The enlargement is 900 - 500 = \boxed{400}.


NP
Sophie walks down the 800 steps of a building, not counting the top step or the ground floor, while Germaine walks up. If Sophie steps on every third step and Germaine steps in the pattern 1, 2, 4, 5, 7, 8, etc., on how many stairs will they both step?
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PostPosted: Wed Sep 16, 2009 4:29 am  Back to top 
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AwesomeToad
Yang-Mills Theory
Yang-Mills Theory


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#3983
Solution
Sophie does not step on 3,6,9, etc. but Germaine does, so they have 0 common steps


New Problem
Evaluate \sqrt{256\cdot32} Express your answer in the form a\sqrt{b} where a and b are integers such that b has no perfect square factors other than 1

PostPosted: Wed Sep 16, 2009 8:06 am  Back to top 
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cgyao15
Navier-Stokes Equations
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#3984
i know it's \sqrt {8192} but how do i simplify it very fast?
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PostPosted: Wed Sep 16, 2009 2:04 pm  Back to top 
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Math Champion
Yang-Mills Theory
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#3985
@cgyao: You don't multiply the two inner numbers first...You first factor the two numbers, then take out all the square factors.
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PostPosted: Wed Sep 16, 2009 3:03 pm  Back to top 
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cgyao15
Navier-Stokes Equations
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#3986
oh thx math champion.

PostPosted: Wed Sep 16, 2009 3:26 pm  Back to top 
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$LaTeX$
Riemann Hypothesis
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#3987
AwesomeToad wrote:

New Problem
Evaluate \sqrt {256\cdot32} Express your answer in the form a\sqrt {b} where a and b are integers such that b has no perfect square factors other than 1


Finishing off the solution, we get \sqrt {256\cdot32}=16\cdot4\sqrt{2}=64\sqrt{4}.

NP: Find the area of the triangle below:
unitsize(15); defaultpen(linewidth(.6pt)+fontsize(4pt)); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((5,0...
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PostPosted: Wed Sep 16, 2009 3:40 pm  Back to top 
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AwesomeToad
Yang-Mills Theory
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#3988
Solution
Pick's Theorem: 6+\frac{13}{2}-2=\frac{21}{2}

NP
Suppose there is a circle with 8 equally spaced points on it. Each is labeled 1,2,3,4,5,6,7, or 8. How many ways are there to label so no 2 labels contain consecutive numbers?
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PostPosted: Thu Sep 17, 2009 3:51 am  Back to top 
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AlphaBetaTheta
Hodge Conjecture
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#3989
hi
that is kinda hard can someone please post the answer please. I have been dying to find out the answer.

PostPosted: Sun Sep 20, 2009 8:10 am  Back to top 
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Math Champion
Yang-Mills Theory
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#3990
I believe you can solve this with Principal of Inclusion/Exclusion
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PostPosted: Tue Sep 22, 2009 2:37 pm  Back to top 
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AwesomeToad
Yang-Mills Theory
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#3991
Re: hi
AlphaBetaTheta wrote:
that is kinda hard can someone please post the answer please. I have been dying to find out the answer.


Hint
Don't label the points yet. How many places can 8 go? Then 7? Then 6? Then 5? etc....


However if someone else has another solution please post it here

PostPosted: Wed Sep 23, 2009 8:28 am  Back to top 
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$AceofDiamonds$
P versus NP
P versus NP

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#3992
This is not that hard first of all.

think of all possible ways that work

and divide out the repeats so

for the first one there are 8 possible ways

the 2nd there are 6 possible ways

so 8*6=48

but we have to divide by 2 giving us an answer of 48/2=24

24

ps: can someone teach me how to do latex.
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PostPosted: Wed Sep 23, 2009 10:51 am  Back to top 
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AwesomeToad
Yang-Mills Theory
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#3993
Please post a new problem Smile

The sum 16^{12}8^{6}2^{5} can be written as 2^N where N is a positive integer.
Find N

PostPosted: Wed Sep 23, 2009 2:16 pm  Back to top 
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$LaTeX$
Riemann Hypothesis
Riemann Hypothesis


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#3994
AwesomeToad wrote:
Please post a new problem Smile

The sum 16^{12}8^{6}2^{5} can be written as 2^N where N is a positive integer.
Find N


Sum? You mean product right? So 16^{12}8^{6}2^{5}=2^{48}2^{18}2^5=2^{71}.

NP: During a severe thunderstorm, Tim saw a flash of lightning and then six seconds later heard a clap of thunder. The speed of sound is 1088 ft/sec. How far is Tim from the lightning? Express your answer to the nearest thousand feet.
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PostPosted: Wed Sep 23, 2009 3:14 pm  Back to top 
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small_unicorn
Poincare Conjecture
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#3995
Quote:
During a severe thunderstorm, Tim saw a flash of lightning and then six seconds later heard a clap of thunder. The speed of sound is 1088 ft/sec. How far is Tim from the lightning? Express your answer to the nearest thousand feet.


Answer 6 000 ft

NP: Abelard and Brendan can do a job together in 10 days. Abelard and Cullen can do it together in 15 days. Brendan and Cullen require 30 days to do the job together. How many days would it take each to do the job if he worked alone ?
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PostPosted: Wed Sep 23, 2009 5:43 pm  Back to top 
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cgyao15
Navier-Stokes Equations
Navier-Stokes Equations


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#3996
We need a
NP

Bob the cow and make 2 gallons on milk every 1/2 hour.
How long will it take to make 4 BUSH(bad bush)els?
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PostPosted: Sun Sep 27, 2009 7:24 am  Back to top 
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goldenboy1.618
Poincare Conjecture
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#3997
Solution
One bushel = 8 gallons, so we need 32 gallons of milk for 4 bushels. Bob makes 4 gal per hour, so this is \boxed{8} hours. (BTW, what is a BUSHel???)


NP
Find the greatest prime factor of 7! + 9!
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PostPosted: Tue Sep 29, 2009 9:55 am  Back to top 
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AwesomeToad
Yang-Mills Theory
Yang-Mills Theory


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#3998
Solution
Since 7!+9!=7!(1+8\cdot9) the largest prime factor is 73

NP
If 3!5!=n! find n!

PostPosted: Wed Sep 30, 2009 8:31 am  Back to top 
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AndrewTom
Navier-Stokes Equations
Navier-Stokes Equations

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#3999
MOEMS Marathon
Solution


n=7



NP


7! \times 6! = 10!. Solve m! \times n! = r! where r > 10.



PostPosted: Wed Sep 30, 2009 10:01 am  Back to top 
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goldenboy1.618
Poincare Conjecture
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#4000
Solution
There exist infinitely many solutions to this problem.
For example, some triples (m, n, r) with m > n are (23, 4, 24), (119, 5, 120), and (719, 6, 720).
The smallest triple that satisfies r > m > n > 10 is (39916799, 11, 39916800).
This triple always exists for (n!-1, n, n!).


BTW, the previous answer is
n = 6 because 6\cdot120 = 720 = 6!.


NP
The sum of a list of positive integers, not necessarily distinct, is 16. Find their maximum possible product.
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PostPosted: Wed Sep 30, 2009 12:28 pm  Back to top 
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