3. Find the largest positive integer $n$ such that $n$ ! ends with exactly 100 zeros.
3. Ans: 409
If $n$ ! ends with exactly 100 zeros, then in the prime factorization of $n$ !, the prime factor 5 occurs exactly 100 times (we need not worry about the prime factor 2 since it will occur more than 100 times). The number of times that 5 occurs in $n!$ is given by
$$
\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{5^{2}}\right\rfloor+\left\lfloor\frac{n}{5^{3}}\right\rfloor+\cdots .
$$

Also the factor 5 occurs 24 times in 100!. Thus the answer is about 400 . Now
$$
\left\lfloor\frac{400}{5}\right\rfloor+\left\lfloor\frac{400}{5^{2}}\right\rfloor+\left\lfloor\frac{400}{5^{3}}\right\rfloor+\cdots=80+16+3=99 .
$$

It follows that 400 ! ends with exactly 99 zeros. Thus the answer is 409 .