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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
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B
-3
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C
none of these
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Solution

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. The minimum value of the sum is -4.


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