Let k be a natural number such that 1 k 20 . Consider a relation R on the set of natural numbers N defined by R = (x, y) : (k^2 - 5)x + (2k + 1)y is a multiple of 11 . The sum of all possible values of k for which R is an equivalence relation is

Step-by-step solution

For R to be an equivalence relation, it must be reflexive. For R to be reflexive, xRx must hold for all x N . (k^2 - 5)x + (2k + 1)x is a multiple of 11 for all x N . (k^2 + 2k - 4)x is a multiple of 11 for all x N . This requires k^2 + 2k - 4 to be a multiple of 11 . k^2 + 2k - 4 0 11 Adding 5 to both sides to complete the square: k^2 + 2k + 1 5 11 (k + 1)^2 5 11 Checking the squares modulo 11 : 1^2 = 1 2^2 = 4 3^2 = 9 4^2 = 16 5 11 5^2 = 25 3 11 Thus, k + 1 4 11 or k + 1 -4 7 11 . This gives k 3 11 or k 6 11 . Fo