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[ Fundamental Theorem of Arithmetic. Prime Factorization]

The number $\frac{100!}{10^{50}}$ was written as an irreducible fraction. Find its denominator.
Since $10^{50}=2^{50} \cdot 5^{50}$, we need to determine which of these one hundred prime factors will cancel out.

Notice that the product 100! contains 50 even factors, meaning that in the prime factorization of 100!, there are at least 50 twos. Therefore, $2^{50}$ will cancel out.

Among the integers from 1 to 100, exactly twenty numbers are divisible by 5 and exactly four of them are divisible by 25. Thus, in the prime factorization of 100!, the number 5 appears with an exponent of 24. After canceling the fraction by $2^{24}$, we get $5^{26}$ in the denominator.

## Answer

$5^{26}$.