B2. What is the largest integer $n$ such that the quantity
$$
\frac{50!}{(5!)^{n}}
$$
is an integer?
Note: Here $k!=1 \times 2 \times 3 \times \cdots \times k$ is the product of all integers from 1 to $k$. For example, $4!=1 \times 2 \times 3 \times 4=24$.
Solution : We can calculate $5!=2^{3} \times 3 \times 5$, thus $(5!)^{n} \mid 50$ ! if
$$
\begin{array}{l}
2^{3 n} \mid 50! \\
3^{n} \mid 50!\text { and } \\
5^{n} \mid 50!
\end{array}
$$
(The notation $b \mid a$ means integer number $a$ is divisible by integer number $b$.)
Observe that among integers from 1 to 50 , every fifth integer is divisible by 5 , and also 25 and 50 are divisible by $5^{2}$. Therefore, the exponent of the maximal power of 5 that divides 50 ! is
$$
\left\lfloor\frac{50}{5}\right\rfloor+\left\lfloor\frac{50}{5^{2}}\right\rfloor=10+2=12
$$
so our desired $n \leq 12$. Doing a similar analysis for the powers of 2 and 3 we get:
$$
\left\lfloor\frac{50}{2}\right\rfloor+\left\lfloor\frac{50}{2^{2}}\right\rfloor+\left\lfloor\frac{50}{2^{3}}\right\rfloor+\left\lfloor\frac{50}{2^{4}}\right\rfloor+\left\lfloor\frac{50}{2^{5}}\right\rfloor=25+12+6+3+1=47,
$$
so $2^{47} \mid 50$ ! and $3 n \leq 47$, thus $n \leq 15$.
$$
\left\lfloor\frac{50}{3}\right\rfloor+\left\lfloor\frac{50}{3^{2}}\right\rfloor+\left\lfloor\frac{50}{3^{3}}\right\rfloor=16+5+1=22,
$$
so $3^{22} \mid 50$ ! and $n \leq 22$.
Since $\min (12,22,15)=12$, the largest integer number $n$ is 12 .
Answer: 12 .