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Problem of the Week #2 - April 9th, 2012

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Chris L T521

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Jan 26, 2012
I would like to thank those of you that participated in the first POTW on MHB. Now it's time for round two! (Bigsmile)

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.

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 $(4123)$].
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Staff member
Jan 26, 2012
Congratulations to the following members for their correct solutions:

1) Sudharaka


Claim: $n$ is even.
Proof: Since each of $a_i's$ is either $1$ or $-1$ each of the summands in the LHS of the given equation is also either $1$ or $-1$. Since the RHS is $ 0$ the number of $1's$ appearing in the LHS is equal to the number of the $-1's$ appearing in the LHS. Thus $2|n$.

Claim: $4|n$.
Proof: From the above claim we can write $n=2k$.
Consider $(a_1a_2a_3a_4)(a_2a_3a_4a_5) \ldots (a_na_1a_2a_3)=(a_1a_2 \ldots a_n)^4$ .
In the above RHS is definitely $1$ while LHS is $(-1)^k (1)^k$.
Thus $(-1)^k(1)^k=1 \Rightarrow (-1)^k=1 \Rightarrow 2|k.$ Hence $4|n$.
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