21. For a positive integer $n$, define $s(n)$ as the smallest positive integer $t$ such that $n$ is a factor of $t$ !. Compute the number of positive integers $n$ for which $s(n)=13$.
21. 792
Solution. For a positive integer $k$, consider the set $A(k)=\{n \in \mathbb{N}: s(n)=k\}$ and we wish to find $\# A(13)$. From the definition of $s(n)$, any element of $A(k)$ must divide $k$ ! but not $(k-1)$ !. Thus, any element of $A(k)$ must be a divisor of $k$ ! that is not a divisor of $(k-1)$ !. We see that $\# A(k)=\sigma_{0}(k!)-\sigma_{0}((k-1)!)$, where $\sigma_{0}(n)$ is the number of divisors of $n$, so that $\# A(13)=\sigma_{0}(13!)-\sigma_{0}(12!)$.
Note that for any prime $p$, the highest power of $p$ that divides $k$ ! is $p^{e}$, where $e=\sum_{j=1}^{\infty}\left\lfloor n / p^{j}\right\rfloor$. Using this formula, we determine the prime factorization of 13 !: 13 ! $=2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$. As $13!=13 \cdot 12$ !, we get $12!=2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7^{1} \cdot 11^{1}$. Hence, we obtain
$$
\# A(13)=11 \cdot 6 \cdot 3 \cdot 2^{3}-11 \cdot 6 \cdot 3 \cdot 2^{2}=198 \cdot 4=792 \text {. }
$$