Question

x! ends with n zeroes. (x+1)! ends with n+2 zeroes. $24\le x\le 626$. How many values are possible for x?

• A. 25
• B. 22
• C. 23
• D. 20

Answer 1

The correct option is D

20

If 24! ends in n zeroes, 25! will end in (n+2) zeroes.
Generalising this to all multiples of 25, they satisfy this condition: if (x)! ends with n zeroes then (x+1)! ends with (n+2) zeroes.
However, if 124! ends in n zeroes, 125! will end in (n+3) zeroes.
Generalising this to all multiples of 125, they do not satisfy the condition: if (x)! ends with n zeroes then (x+1)! ends with (n+2) zeroes.
Infact, they end with (n+3) or (n+4) zeroes.
Therefore, the number of favourable cases between 24 to 626 can be represented as:
(Number of multiples of 25 - Number of multiples of 125)

25 - 5 = 20.