Question

Given that $p,q,r,s,t,u$ are integers and $p+q+r+s+t+u=2005$, what is the minimum value of $(-1{)}^p+(-1{)}^q+(-1{)}^r+(-1{)}^s+(-1{)}^t+(-1{)}^u?$

• A. -2
• B. -3
• C. none of these

Answer 1

The correct option is C none of these

$p+q+r+s+t+u=2005$. Of the 6 integers, the number of integers that can be odd is 5, 3 or 1.

In the expression $(-1{)}^p+(-1{)}^q+(-1{)}^r+(-1{)}^s+(-1{)}^t+(-1{)}^u$:

1) When one integer is even and the others are odd, then only one among is +1 and others are -1 each. Hence the sum of the terms is -5 + 1 = -4.

2) When three of the integers are even, then three of the terms are +1 each and the remaining three are -1 each. Hence the sum is 0.

3) When five of the integers are even, then five terms will be +1 each and the other term is -1. Hence the sum is 4.

$\therefore$ The minimum value of the sum is -4.