Question

The number of zeros at the end of $60!$ is:

• A. $12$
• B. $14$
• C. $16$
• D. $18$

Answer 1

Answer: (B)

$14$

The number of trailing zero in $n!$

$=\frac{n}{5}+\left[\frac{n}{5^2}\right]+......+\left[\frac{n}{5^k}\right]\left[\right]⟶G.I.F$

Where $5^{k+1}>n$

Number of zero in $60!=?$

$n=60=\frac{n}{5}+\left[\frac{n}{5^2}\right]=\frac{60}{5}+\frac{60}{25}=12+2=14$

$\therefore 14$ Zeros are there