Question

The largest value of $n$, so that ${10}^n$ divides $51!$

• A. 47
• B. 12
• C. 35
• D. 59

Answer 1

The correct option is B 12

$i.e.[\frac{51}{2}]+[\frac{51}{2^2}]+[\frac{51}{2^3}]+[\frac{51}{2^4}]+[\frac{51}{2^5}]=[25.5]+[12.75]+[6.375]+[3.1875]=25+12+6+3=46$
$\left[\frac{51}{5}\right]+\left[\frac{51}{5^2}\right]=12$
$\Rightarrow 51!=2^{46}{.3}^{\alpha }{.5}^{12}{.7}^{\beta }\dots \dots$
$\Rightarrow 51!$ ends with $2^{12}{.5}^{12}={10}^{12}$. i.e. $12$ zeros

$\therefore$ Largest value of n is $12$