## Saturday, 5 November 2011

### Numbers-13 (XAT-2009)

F(x) is a fourth order polynomial with integer coefficients and no common factor. The roots of F(x) are -2,-1,1,2. If p is prime number greater than 97, then the largest integer that divides F(p) for all values of p is:
A.72                B. 120             C. 240             D. 360             E. None of the above
Solution follows here:
Solution:
The roots of F(x) are -2,-1,1,2 => F(x) = (x+2)(x+1)(x-1)(x-2)
F(p) = (p+2)(p+1)(p-1)(p-2) = (p+2)(p+1)(p)(p-1)(p-2)/p
=> p*F(p) = (p+2)(p+1)(p-1)(p-2) = (p+2)(p+1)(p)(p-1)(p-2)
p*F(p) is a product of 5 consecutive numbers
=> it is divisible by 5    ---(1)
Middle number p is prime => obviously it is odd => (p-1) and (p+1) are consecutive even numbers => For two consecutive even numbers, one must be a multiple of 4 and the other must be a multiple of 2  => p*F(p) is divisible by 4*2 ie., 8       ---(2)
One of the three consecutive numbers (p+2),(p+1),p is a multiple of 3 => as p is prime, either (p+2) or (p+1) is a multiple of 3 => similarly, either, (p-2) or (p-1) is a multiple of 3 =>  p*F(p) is divisible by 3*3 ie., 9                                                                      ---(3)
From (1),(2) and (3), p*F(p) is divisible by 5*8*9 ie., 360
=> As p is prime, F(p) is divisible by 360
Note:
As it is given that p > 97, it should not be one of 2 or 3 or 5 and hence we can surely say that, "As p*F(p) is divisible by 360F(p) is divisible by 360". If not, p can be one of 2 or 3 or 5, then F(p) can be divisible by a 360/2 or 360/3 or 360/5 ie., 180 or 120 or 72.
Answer (D)