Another well attempted problem with solutions from Mary of Birchwood High School, Sana, Jenny, Chris and Rosion of Madras College, St. Andrews, as well as Andrei of School 205 Bucharest, Chen of the Chinese High School, Singapore.

100! = 100 x 99 x 98 x 97 x 97 x ....... x 4 x 3 x 2 x 1

To write this as the product of its prime factors:

Considering some examples:
100 = 22 x 52
80 = 24 x 5

So to find all the 2's that appear as prime factors it is necessary to consider all the numbers divisible by powers of 2 between 1 and 100
Number of numbers divisible by 2 = 21 is50 (the even numbers - with at least one 2 as a factor)
Number of numbers divisible by 4 = 22 is 25 (with at least two 2's as factors)
Number of numbers divisible by 8 = 23 is 12 (with at least three 2's as factors)
Number of numbers divisible by 16 = 24 is 6 (with at least four 2's as factors)
Number of numbers divisible by 32 = 25 is 3 (with at least five 2's as factors)
Number of numbers divisible by 64 = 26 is 1(with six 2's as factors).

Therefore number of 2's is 50 + 25 + 12 + 6 + 3 + 1 = 97

Similarly:
Number of numbers divisible by 5 = 51 is 20 (the even numbers - with at least one 2 as a factor)
Number of numbers divisible by 25 = 52 is 4 (with at least two 2's as factors)

Therefore number of 5's is 20 + 2 = 24