\section*{Problem 4 - 031044}

How many trailing zeros does the product

\[
p_{1}^{1} \cdot\left(p_{1}^{2} \cdot p_{2}^{1}\right) \cdot\left(p_{1}^{3} \cdot p_{2}^{2} \cdot p_{3}^{1}\right) \cdot \ldots \cdot\left(p_{1}^{100} \cdot p_{2}^{99} \cdot p_{3}^{98} \cdot \ldots p_{98}^{3} \cdot p_{99}^{2} \cdot p_{100}^{1}\right)
\]

have, where \(p_{1}, p_{2}, p_{3}, \ldots, p_{100}\) are the first hundred prime numbers in their natural order.
}

The number of trailing zeros depends on the number of factors \(p_{1}=2\) and \(p_{3}=5\); because only the product of the two prime numbers 2 and 5 yields a zero.

In the given product, the number 2 appears exactly \((1+2+3+\ldots+100)\) times, and the number 5 appears exactly \((1+2+\) \(3+\ldots+98)\) times as a factor.

Thus, the number has exactly \(1+2+3+\ldots+98=98 \cdot \frac{99}{2}=4851\) trailing zeros.

Taken from \([5]\)