1239! Ends With Many Zeros?

If a number has $\textit{n}$ terminal $\textit{zeros}$ , then it contains $\textit{n}$ factors of $10$ . But since $10=5\cdot 2$ we are asking how many how many factors of $2$ and $5$ are there.

Since $2$ is smaller than $5$ , for any factor of $5$ there will be enough factors of $2$ to make a factor of $10$ . thus we need to count only factors of $5$

Since $1239 \div 5 = 247.8$

there are $247$ factors of $5$ on $1239$

But the numbers divisible by $25$ will have $2$ factors of $5$ of which $1$ we have counted.

Since, $1239 \div 25 = 49.56$

there are $49$ factors of $5$ again

Similarly, there will be $3$ factors of $125$ in $1239$ of which $2$ we have counted

Since, $1239 \div 125 = 9.912$

there are $9$ factors of $5$ again

similarly there will be $1$ factor again of $5$ taking into consideration $625$ which contain $4$ factors of $5$

Therefore, there are total $247+49+9+1 = 306$ factors of $5$

which implies that there are total $306$ $zeros$ at the end of $1239!$

There are: 247 numbers divisible by 5; 49 numbers divisible by 25; 9 numbers divisible by 125; 1 number divisible by 625; Hence there are: 247+49+9+1=306 numbers 0.