Let S be the set of all integers x satisfying the inequality x^2 - 120x + 2000 0 . A relation R is defined on the set of integers such that aRb if and only if gcd (a, b) > 1 . If A = x S : xR42 , then the sum of all elements in the set A is _____.

Step-by-step solution

First, solve the quadratic inequality to find the set S : x^2 - 120x + 2000 0 (x - 20)(x - 100) 0 Thus, x [20, 100] . Since x is an integer, S = 20, 21, 22, , 100 . The set A consists of elements x S such that xR42 , which means gcd (x, 42) > 1 . Since 42 = 2 3 7 , an element x S belongs to A if it is divisible by 2 , 3 , or 7 . We need to find the sum of all such elements. Let S_k denote the sum of multiples of k in the set S . By the Principle of Inclusion-Exclusion: Sum (A) = S₂ + S₃ + S₇ - S₆ - S₁₄ - S₂₁ + S₄₂