Problem 9.1. For a natural number $a$, the product $1 \cdot 2 \cdot 3 \cdot \ldots \cdot a$ is denoted as $a$ !.

(a) (2 points) Find the smallest natural number $m$ such that $m$ ! is divisible by $23 m$.

(b) (2 points) Find the smallest natural number $n$ such that $n$ ! is divisible by $33n$.
# Answer:

(a) (2 points) 24.

(b) (2 points) 12.

Solution. (a) The condition is equivalent to $(m-1)!$ being divisible by 23. Since 23 is a prime number, at least one of the numbers $1, 2, \ldots, m-1$ must be divisible by 23, so $m-1 \geqslant 23$ and $m \geqslant 24$. Clearly, $m=24$ works, since in this case $\frac{24!}{23 \cdot 24}=22!$.

(b) The condition is equivalent to $(n-1)!$ being divisible by $33=3 \cdot 11$. Since 3 and 11 are prime numbers, at least one of the numbers $1, 2, \ldots, n-1$ must be divisible by 11, so $n-1 \geqslant 11$ and $n \geqslant 12$. Clearly, $n=12$ works, since in this case $\frac{12!}{33 \cdot 12}=\frac{10!}{3}=8! \cdot 3 \cdot 10$.