Mathcounts marathon!!!!

AwesomeToad

NP

**Posted:** Thu Nov 05, 2009 9:18 am

isabella2296

We can factor out a $7^{45}$ from the top to get $\frac{7^{45}(49 - 7 - 1)}{41} \implies \frac{7^{45}(41)}{41}$ . The 41's cancel out, leaving us with $7^{45} = 7^{x}$ , so $x = 45$ .

New problem:

Andy has 3 identical peppermints and 2 identical candy canes. How many ways can he arrange the 5 candies on a table?

**Posted:** Thu Nov 05, 2009 9:23 am

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Andy can arrange them $\frac{5!}{2!3!}=\boxed{10}$ ways.

NP: The set $\{5, 8, 10, 18, 19, 28, 30, x\}$ has eight members. The
arithmetic mean of the set's members is 4.5 less than x. What is
the value of x ?

**Posted:** Thu Nov 05, 2009 2:41 pm

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steve123456

(5 + 8 + 10 + 18 + 19 + 28 + 30 + x)/8=x-4.5. 5 + 8 + 10 + 18 + 19 + 28 + 30 + x =8x-36.x= 8x- 154. 7x=154, so x=22

NP: What value of x satisfies the equation 25^-2 = (5^48/x)/(5^[26/x] * 25^[17/x]) Sorry I can't use Latex.

**Posted:** Thu Nov 05, 2009 10:32 pm

AwesomeToad

**Posted:** Fri Nov 06, 2009 9:13 am

fishythefish

And add a few braces...and fractions if you want:

$25^{-2}=\frac{(5^{\frac{48}{x}})}{(5^{\frac{26}{x}})(25^{\frac{17}{x}})}$

$5^{-4}=5^{\frac{5}{x}}$

$-4=\frac{5}{x}$

$x=-\frac{5}{4}$

NP:

In one of those "What color hats are the prisoners wearing?" logic problems, there are 10 (distinguishable) prisoners, and 1 red hat, 2 yellow hats, 3 green hats, and 4 blue hats. How many different orders of hats can there be when the prisoners are lined up in order? (Note: hats of the same color are not distinguishable.)

**Posted:** Sat Nov 07, 2009 2:38 pm

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Solution
NP

**Posted:** Sat Nov 07, 2009 2:48 pm

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fishythefish

Should we assume that you are just adding 4 each time? (There are multiple ways to write/solve this, but for simplicity's sake, I'll start the arithmetic series at $a_1=37$ .)

$\displaystyle\sum_{n=1}^{17}(33+4n)$

$S_{17}=\frac{17(37+101)}{2}=\frac{17(138)}{2}=\frac{2346}{2}=1173$

NP: What is the answer to my previous problem if there are 5 black prisoners and 5 white ones? (And those of the same color are not distinguishable.)