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CAT 2000 — QA Question 15

Even OddEasy

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,

Answer & solution

  • n is even

  • B

    n is odd

  • C

    n is an odd multiple of 3

  • D

    n is prime

Solution

The terms in the given expression are x1,  x2, ... xn-1, xn.

There are n terms in the expression.

The only possible value of each of the terms is either 1 or –1.

For the given expression to be zero there should be even number of terms of which half the terms are –1 and the remaining are 1.

∴ n is even.

Hence, option (a).

CAT 2000 QA Q15: Each of the numbers x 1 , x 2 ...., x n , n > 4, is equal to 1 or –1. Suppose, x 1 x 2 x 3 x 4 + x 2 x 3 — Solution | TheCATExam