The problem is as follows
For every positive even integer n, the function h(n) is defined to be the product of all the 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
h(100)=the product of all even integers from 2 to 100
There is one golden rule u need to remember. The product of n consecutive multiples of a certain number is always divisible by n! and therefore n Now all the even numbers from 2-100 are nothing but multiples of 2 so this golden rule applies here. Firstly, we need to find how many multiples of 2 are there betn 2-100 inclusive. (100-2)/2+1=50. So, h(100)=z(50!) where z is some integer.
Now, the largest prime factor that z(50!) has is the number 47. So when 1 is added to z(50!) the resultant number is not going to be a multiple of 47. Because when one multiple is added/subtracted to another non multiple the result is ALWAYS a non-multiple. The same can be said of all the other prime factors of z(50)! eg 43,41 etc. So the smallest multiple of z(50!) has to be greater than 47. But 48,49 are not primes so the smalles prime divisor of z(50!)has to be greater than 50 Hence. E
