One solution of the equation
(x-a)(x-b)(x-c)(x-d)=25
is x=7.

If a, b, c, and d are different integers, find the value of a+b+c+d

Since x=7 is a solution of the given equation, it follows that:

P*Q*R*S = 25, where (P, Q, R, S) = (7-a, 7-b, 7-c, 7-d)(say) ...(#)

Since a, b, c, d are distinct integers, so are P, Q, R, S, and accordingly, we must have:

(P, Q, R, S) (- (5, -5, 1, -1). In other words, (P, Q, R, S) corresponds to a perm of (5, -5, 1, -1) in consonance with (#). No other factorization of 25 will yield distinct integer values for the quadruplet (P, Q, R, S).

Thus, P+Q+R+S = 5-5+1-1 = 0, giving:

28-(a+b+c+d) = 0, so that:

a+b+c+d = 28

Consequently, the required value of a+b+c+d is 28.

*Edited on ***July 10, 2008, 6:42 am**