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Home » Mathematics » Middle School Contests (MATHCOUNTS, AMC 8, MOEMS, Ages 10-13)
 
Middle School Marathon
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BuckeyeFan100
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#41
FlyAgaric wrote:
brainomega wrote:
18. How many ways could the Pentagon's walls be painted with 8 colors if no two adjacent sides may have the same color?


Do flips count?


I am assuming not.

There would be 8*7*7*7*6 = 16464 \text{ways}

Next Problem is Maybachs
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PostPosted: Sun Jul 12, 2009 2:23 pm  Back to top 
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AIME15
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#42
That is wrong. For example, ABCA(x) where (x) denotes some other color is not counted correctly.
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PostPosted: Sun Jul 12, 2009 2:32 pm  Back to top 
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AIME15
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#43
[hide=Solution to maybach's problem]
Ron gains 15 miles on him each hour - thus, the answer is 15 \times 2=\boxed{30}.
[/hide]

Current Problem:

How many ways could the Pentagon's walls be painted with 8 colors if no two adjacent sides may have the same color?

I think we can use recursion for this...?
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PostPosted: Mon Jul 13, 2009 11:48 am  Back to top 
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Math Champion
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#44
By the Pentagon, do you mean the U.S. defense one? Cause that one has three pairs of inside and outside walls. Or do you just mean a regular Pentagon?
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PostPosted: Mon Jul 13, 2009 4:46 pm  Back to top 
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cgyao15
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#45
yes, he put Pentagon in upper case, so it's the Pentagon=pwnage of knowledge.

PostPosted: Mon Jul 13, 2009 4:47 pm  Back to top 
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#46
Then this problem requires a LOT, and I mean LOT, of casework.
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PostPosted: Mon Jul 13, 2009 4:50 pm  Back to top 
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pytheagle
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#47
I'll post a new problem to keep this marathon going.

Problem: Find a+b when: 2009a+2008b=20072008a+2009b=7002
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PostPosted: Mon Jul 13, 2009 5:54 pm  Back to top 
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brainomega
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#48
Quote:
18. How many ways could the Pentagon's walls be painted with 8 colors if no two adjacent sides may have the same color?


Exterior walls (not that it's much harder to find all of the walls)

PostPosted: Mon Jul 13, 2009 9:14 pm  Back to top 
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mismatchtea
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#49
Quote:

2009a + 2008b = 2007

2008a + 2009b = 7002


Adding the two equations:
4017a + 4017b = 9009
a + b = 9009/4017

Now, for another problem:

Find the sum of n terms of: 2/3 + 8/9 + 26/27 + ...

PostPosted: Tue Jul 14, 2009 2:34 am  Back to top 
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picture74
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#50
mismatchtea wrote:

Now, for another problem:

Find the sum of n terms of: 2/3 + 8/9 + 26/27 + ...


Do you mean 2/3 + 8/9 + 16/27 + ... Question

PostPosted: Tue Jul 14, 2009 3:23 am  Back to top 
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mismatchtea
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#51
Quote:

Do you mean 2/3 + 8/9 + 16/27 + ... Question


No, I do not.

For more terms:
2/3 + 8/9 + 26/27 + 80/81 + 242/243 + ...

Hopefully, the pattern is more clear now.

PostPosted: Tue Jul 14, 2009 3:44 am  Back to top 
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picture74
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#52
mismatchtea wrote:

For more terms:
2/3 + 8/9 + 26/27 + 80/81 + 242/243 + ...



General term t_n=\frac{3^n-1}{3^n}

Sum=\sum_{n=1}^{n} \frac{3^n-1}{3^n}

=\sum_{n=1}^{n} 1 - \sum_{n=1}^{n} \frac{1}{3^n}

=n-\frac{1(1-(\frac{1}{3})^n)}{3(1-\frac{1}{3})}

=n-\frac{3^n-1}{2 \times 3^n}

I don't have a new problem presently, so someone else may post one

PostPosted: Tue Jul 14, 2009 4:35 am  Back to top 
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Mukherjee
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#53
In how many ways can 5 subjects be arranged in 6 periods of a class?

PostPosted: Tue Jul 14, 2009 4:57 am  Back to top 
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SuperNerd123
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#54
Can the subjects go on for more than 1 period?

PostPosted: Tue Jul 14, 2009 6:28 am  Back to top 
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Mukherjee
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#55
Yes.They can.

PostPosted: Tue Jul 14, 2009 9:27 am  Back to top 
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PowerOfPi
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#56
Is it just 5!=120?
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PostPosted: Tue Jul 14, 2009 11:27 am  Back to top 
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isabella2296
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#57
Wouldn't it be 6 \times 5 \times 4 \times 3 \times 2 = 720?
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PostPosted: Tue Jul 14, 2009 12:25 pm  Back to top 
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brainomega
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#58
I think it's 5*5*4!=600 arrangements. There are 5 ways to choose which class is two periods, 5 ways to place that class, and 4! ways to place the remaining classes. If this is correct, someone can post a new problem.

PostPosted: Tue Jul 14, 2009 12:40 pm  Back to top 
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pytheagle
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#59
Why didn't you yourself post a new problem?

Problem: Find the harmonic mean of the first 3 composite numbers.
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PostPosted: Tue Jul 14, 2009 12:53 pm  Back to top 
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brainomega
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#60
solution
The first 3 composite numbers are 4,6, and 8. \frac1{\frac14+\frac16+\frac18}=\frac{24}{13}


np=pentagon problem near top of page

PostPosted: Tue Jul 14, 2009 1:02 pm  Back to top 
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