Question

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

Answer 1

x! ends with n zeroes
(x + 1)! ends with n + 2 zeroes
$24\le x\le 626$
as x! ends with n zeroes
So it will be have $(5\times \mathrm{5...}ntimes)(2\times 2\times ....mtimes)$
$=\text{/}{5}^n\text{/}{5}^m{5}^n{2}^m(m>n)$
(x + 1)! ends with (n + 2) zeroes
$\therefore (x+5)!$ will have $5^{n+2}{2}^m(m>n)$
So, we need to find number of multiples of 25 but not multiples of 125
As if it is 125, then there will be one more 5 which give one more zero
So between 24 - 626
Theri will be 25 number of multiples of 25 but out those 25, 5 numbers i.e. 125, 250, 315, 500 and 625 (Which are multiples of 125) so we will substract them.
$\therefore$ there will be 20 possible values of x.