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?
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.