To find the number of 2's , 3's or any prime number in a factorial say N!, we keep dividing N (not the factorial value of N, but only N) by the prime number. If the quotient obtained is greater than or equal to the divisor, we keep dividing each quotient by the prime number until we get a value of the quotient less than the divisor. The sum of all the quotients gives us the required value
For eg. how many 3's are there in 100!
Then 100 ÷ 3 = 33 (we need the quotient only, not the remainder.)
Since 33 is > 3, then 33 ÷ 3 = 11
Since 11 is > 3, then 11 ÷ 3 = 3
Since 3 = 3, then 3 ÷ 3 = 1
Therefore the number of 3's in 100! = 33 + 11 + 3 + 1 = 48
This method becomes very important when we need to find the number of trailing 0's in a factorial. In that case the divisor is 5. For eg how many trailing 0's are there in 1000!
1000 ÷ 5 = 200
240 ÷ 5 = 48
48 ÷ 5 = 9
9 ÷ 5 = 1
Therefore there are 249 0's at the end of 1000!.
________________________________________
In the question, we need to find the number of 2's in 100!
100 ÷ 2 = 50
50 ÷ 2 = 25
25 ÷ 2 = 12
12 ÷ 2 = 6
6 ÷ 2 = 3
3 ÷ 2 = 1
Therefore the number of 2's = 50 + 25 + 12 + 6 + 3 + 1 = 97
Option A
Arun Kumar