When the number

$$
N=1^{1} \times 2^{2} \times 3^{3} \times \cdots \times 9^{9}
$$

is written as a decimal number, how many zeros does it end in?
The number $N$ ends with five zeros. Indeed, $10^{5}$ divides $N$ since $4^{4} \times 5^{5}=$ 800,000 divides $N$. But $10^{6}$ does not divide $N$, since the only terms in the product which are divisible by 5 are $5^{5}$, and hence $5^{6}$ does not divide $N$. So the answer is 5 .