\section*{Problem 5 - V11035}

The number \(n\)!, read as " \(n\) factorial", refers to the product \(1 \cdot 2 \cdot 3 \cdot \ldots \cdot (n-1) \cdot n\) of all natural numbers from 1 to \(n\).

For example, \(4! = 1 \cdot 2 \cdot 3 \cdot 4 = 24\).

How many trailing zeros does the number 50! (50 factorial) have? Justify your answer!
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We decompose all factors into their prime factors. An end-zero is created precisely when the prime numbers 2 and 5 are multiplied together. The number of fives is determined by the factors that have a zero or a five as their last digit, i.e.:

\[
5, 10=2 \cdot 5, 15=3 \cdot 5, 20=2^{2} \cdot 5, 25=5^{2}, 30=2 \cdot 3 \cdot 5, 35=5 \cdot 7, 40=2^{3} \cdot 5, 45=3^{2} \cdot 5, 50=2 \cdot 5^{2}
\]

In total, the prime factor 5 appears 12 times. Since the prime factor 2 appears in every even number, the number 50! has a total of 12 trailing zeros.