Problem Solving

For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is

(a) between 2 and 10
(b) between 10 and 20
(c) between 20 and 30
(d) between 30 and 40
(e) greater than 40

OA is E

How to solve this?

By definition:
H(100) = (2*4*6*.....*98 *100) ----(50 even terms within the parenthesis)
= > H(100) = 2^50(1*2*3*4....*49*50) ----(factoring 2 from each of the 50 even terms)
Thus H(100) is divisible by all primes between 1 & 50
=> H(100)+1 = 2^50(1*2*3*4....*49*50) + 1
1 is not evenly divisible by any of the primes between 1 & 50 => the least prime to divide H(100) has to be greater than 50 (In terms of primes, it must be greater than the highest prime below 50 which is 47)
{Note: For any number n = a*b*c, where a, b, c are positive primes greater than 1, if you add 1 to n, the resulting (n+1) cannot be divided by a or b or c.}

So for H(100)+1 = 2^50(1*2*3*4.......*49*50) + 1
=> by adding 1 the expression [H(100)+1] becomes indivisible by any of the primes in "1*2*3*4.......*49*50"
=> because of the nature of the answer choices, it is sufficient for us to know that the no prime under 50 is a factor of H(100)+1
We do not need to test for the value of factors of H(100)+1 beyond that.
Hence [spoiler]E: "greater than 40"[/spoiler]
Hope this helps...

**Please note that the prime factors of h(n)+1 will be greater than the largest prime factor of h(n). As any factor of h(n) when divides H(n)+1 will give a remainder of 1.

So, h(100)=100*98*96....*2
=2^50(50*49*48.....*1)

We can see that any number between 1 to 50(including all the primes) will be a factor of h(n).

So we can conclude in conjunction with above property** that the smallest prime factor will be greater than 50 which seems to coincide with option E