Quote:
Originally Posted by dissa
hmm.. b/n between? I didn't get it. 
My answer is here,
h(n) = 2 x 4 x .....n
h(100) = 2 x 4 x ...100
h(100) + 1 = (2 x 4 x ...100 + 1) = 2^50 (1 x 2 x...50) +1
if we divide this any number below 50;
1 would be the remainder. So that the smallest prime factor would be greater than 50.
answer (E) |
I will elaborate on this more to make it simple.
h(n) = 2 x 4 x .....n
h(100) = 2 x 4 x ...100
Since there are 50 even nos. in 1-100 we can write:
h(100) = 2^50 *(1x2x3x4...50)
now, h(100) will be divisible by each nos. from 1 to 50 as it is in the numerator.
but, h(100) + 1 = (2 x 4 x ...100 + 1) = 2^50 (1 x 2 x...50) +1
with 1 added to it these prime nos. cannot divide the nos. in simpler terms if A can divide B, A cannot divide B+1. say 2 can divide 10, but cannot divide 11.
Coming back to the problem, since h(100) is divisible by all prime nos. from 1-50, the no. which can divide h(100)+1 has to be greater than 50.
Hence, the answer is E