Each of the numbers \(x_1,x_2,....,x_n\) equal \(1\) or \(-1\) and \[x_1x_2x_3x_4 + x_2x_3x_4x_5 + x_3x_4x_5x_6+...+x_{n-1}x_nx_1x_2 + x_nx_1x_2x_3 = 0\]

Prove that \(n\) is divisible by \(4\).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestIn this proof the number of \(-1\) in an expression means the number of variables \(x_i\) that are equal to -1. Since each variable \(x_i\) appears four times in the expression, the number of \(-1\) on the LHS is a multiple of 4 or even($). Since each term is either 1 or -1 and they sum to 0, half of them must be -1(this implies n is even) and the other half 1. A term is equal to -1(1) \(\iff\) it has an odd(even) number of \(-1\), therefore there must be an even number of terms with an odd number of \(-1\) to make statement ($) true, this means \(\frac {n}{2}=\)even so n is a multiple of 4.

Log in to reply

Nice!

Log in to reply

If we multiply them all together, we will get 1.So the number if \(-1\)'s of the form \( x_i.x_j.x_q.x_r\) is even.But since the sum of all the terms is 0, then the \(-1\)'s will be half of all of them, which are n in total.Thus n is divisible by 4.

Log in to reply

Your idea is correct but you should explain it a bit more in my opinion because anyone who is not used to this type of problem may find it hard to decipher the solution .......

Log in to reply