Each of the numbers x1, x2...., xn, n > 4, is equal to 1 or –1.
Suppose, x1x2x3x4 + x2x3x4x5 + x3x4x5x6 + ... + xn–3xn–2xn–1xn + xn–2xn–1xnx1 + xn–1xnx1x2 + xnx1x2x3= 0, then,
Options:
A) n is even  B) n is odd.  C) n is an odd multiple of 3.  D) n is prime  E) None
Explanation :   Every term of the eq : x1x2x3x4 + x2x3x4x5 + x3x4x5x6 + ... + xn–3xn–2xn–1xn + xn–2xn–1xnx1 + xn–1xnx1x2 + xnx1x2x3= 0, would either be 1 or -1 (product of 1's and -1's would never be anything else).   Now for the equation to be true number of 1's would have to be equal to number -1's thus there would always be even number of terms in the equation. Now for that to be true n would have to be even.   Say for example n=6: x1x2x3x4 + x2x3x4x5 + x3x4x5x6 + x6x1x2x3.   We see there are 4 terms in the equation, we can have 2 1's and 2 -1's. So , n is even.   n is evennswer : n is even
<result>A</result>
Each of the numbers x1, x2...., xn, n > 4, is equal to 1 or –1.
Suppose, x1x2x3x4 + x2x3x4x5 + x3x4x5x6 + ... + xn–3xn–2xn–1xn + xn–2xn–1xnx1 + xn–1xnx1x2 + xnx1x2x3= 0, then,