#Numbers - 18
F(x) is a fourth order polynomial with integer coefficients and with no common factor. The roots of F(x) are -2,-1,1,2.If P is a prime number greater than 97,then the largest integer that divides F(p) for all the values of p is (a)72 (b)120 (c)240 (d)360 (e)None of these Solution - F(x) = (x-2)(x-1)(x+1)(x+2) P must be of the form - 6k+1/6k-1 Substituting the value of p(6k-1) in the eqn, (6k-3)(6k-2)(6k)(6k+1) 3*2*6=36 Substituting the value of p(6k+1) in the eqn, (6k-1)(6k)(6k+2)(6k+3) 6*2*3=36 so 72 must be a factor now options (a),(d) and (e) are left, now they are 5 consecutive numbers with middle term missing, starting with an odd number (6k-1)(6k+2)(6k)(6k+3)or (6k-3)(6k-2)(6k)(6k+1) the number will always contain a multiple of 5 36*2*5=360 ANS