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- Jan 26, 2012

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I would like to thank those of you that participated in the first POTW on MHB. Now it's time for round two!

This week's problem was proposed by caffeinemachine.

There were no hints provided for this problem.

Remember to read the POTW submission guidlines to find out how to submit your answers!

Here are some more terms in the sum: $a_1a_2a_3a_4 + a_2a_3a_4a_5 + a_3a_4a_5a_6+a_4a_5a_6a_7+\cdots+a_{n-2}a_{n-1}a_na_1 + a_{n-1}a_na_1a_2+a_na_1a_2a_3$.

You may now be asking what happens in the case of $n=4$. Well, if you follow this pattern, you are cycling through the indices 4 at a time, and once you get to $n$, you "start over again." So, if $n=4$, the indices of the terms in the sum will be a cyclic permutation of $(1234)$ [starting with $(1234)$ and ending with $(4123)$].

This week's problem was proposed by caffeinemachine.

**Problem**: Let each of the numbers $a_1, a_2, \ldots , a_n$ equal $1$ or $-1$. If we're given that $a_1 a_2 a_3 a_4 + a_2 a_3 a_4 a_5 + \ldots + a_n a_1 a_2 a_3 =0$, prove that $4|n$.There were no hints provided for this problem.

Remember to read the POTW submission guidlines to find out how to submit your answers!

**EDIT**: Since there was some confusion as to what this was asking, I included a clarification in the spoiler.You may now be asking what happens in the case of $n=4$. Well, if you follow this pattern, you are cycling through the indices 4 at a time, and once you get to $n$, you "start over again." So, if $n=4$, the indices of the terms in the sum will be a cyclic permutation of $(1234)$ [starting with $(1234)$ and ending with $(4123)$].

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