【Question 23】
How many consecutive "0"s are at the end of the product $5 \times 10 \times 15 \times 20 \times \cdots \times 2010 \times 2015$?
【Analysis and Solution】
Number Theory, Prime Factorization, Trailing Zeros.
We know that if the product of some numbers ends with zero, it means that the prime factorization of these numbers must include the prime numbers 2 and 5.
Each additional pair of 2 and 5 results in one more zero.
$$
\begin{array}{l}
5 \times 10 \times 15 \times 20 \times \cdots \times 2010 \times 2015 \\
=(5 \times 1) \times(5 \times 2) \times(5 \times 3) \times(5 \times 4) \times \cdots \times(5 \times 402) \times 5 \times(403) \\
=(1 \times 2 \times 3 \times 4 \times \cdots \times 402 \times 403) \times 5^{403} ;
\end{array}
$$

We consider the number of factors of 2 and 5 in the product \(1 \times 2 \times 3 \times 4 \times \cdots \times 402 \times 403\);
$$
\begin{array}{l}
403 \div 2=201 \cdots \cdots 1, \quad 201 \div 2=100 \cdots \cdots 1, \quad 100 \div 2=50,50 \div 2=25, \\
25 \div 2=12 \cdots \cdots 1, \quad 12 \div 2=6, \quad 6 \div 2=3, \quad 3 \div 2=1 \cdots \cdots 1 ; \\
201+100+50+25+12+6+3+1=398 ;
\end{array}
$$
The product \(1 \times 2 \times 3 \times 4 \times \cdots \times 402 \times 403\) contains 398 factors of 2;
Therefore, the product \(5 \times 10 \times 15 \times 20 \times \cdots \times 2010 \times 2015\) contains 398 factors of 2;
While the product \(5 \times 10 \times 15 \times 20 \times \cdots \times 2010 \times 2015\) contains more than 403 factors of 5;
Hence, the product \(5 \times 10 \times 15 \times 20 \times \cdots \times 2010 \times 2015\) has 398 consecutive "0"s at the end.