According the given information h(100)= 2*4*6*8*10......*100
If we take 2 as common from each of those even number we get h(100)= 2^50 (1*2*3*4*.....*50).
Now, Consider any number, eg. 30 = 1*2*3*5 the prime factors over here are 2,3,5. But if we add 1 to 30 the new number (in this case 31) will not be divisible by any of 30's prime factor i.e. 2,3,5 because though 30 is divisible by 2,3,5 .. 1 is not divisible by either of the three prime factors 2,3,5.
For h(100) the prime factors are all the prime numbers between 1 and 50. Now, when we add 1 to h(100) the resulting number will not be divisible by any of the prime factors of h(100) (the same reason as in the above example)and the smallest prime factor will have to be greater than 49.
