2 consecutive integers are always relatively prime. E.g 4 & 5, or 9 & 10 etc.
If two numbers are relatively prime, then factor of one number leaving 1 cannot be factor of other number. E.g 9 has factor 3 but 10 does not have it as factor and 10 has factor 5, which is not shared by 9.
This question focuses on the above mentioned concept.
h(100) can be written as = 2.4.8.16...100 = 2*1.2*2.2*3.2*4....2*50 = 2^50(1.2.3....50).
h(100)+1 will be immediate number to h(100), so it will be relatively prime number to h(100).
As h(100) has all the factors from 1 to 47 (biggest prime number from 1 to 50, I did not consider 48, 49, and 50 as they can be reduced to smaller prime numbers).
So factor of h(100) + 1 will have factor which is at least greater than 47.
Answer E.