#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
(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
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