1. A natural number $m>1$ is called a $k$-fold divisor of a natural number $n$ if the equality $n=m^{k} q$ holds, where $q$ is an integer that is not a multiple of the number $m$. Determine how many sevenfold divisors the number $100 !=1 \cdot 2 \cdot 3 \cdot \ldots \cdot 100$ has.
Solution. First, we derive a general formula for the number $P_{k}(n)$ of all $k$-fold divisors of a number $n$ with the factorization $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \ldots p_{N}^{a_{N}}$, where $p_{i}$ are distinct primes and the exponents $a_{i}$ are non-negative integers. It holds that $m^{k} \mid n$ if and only if $m$ is of the form $p_{1}^{b_{1}} p_{2}^{b_{2}} \ldots p_{N}^{b_{N}}$, where the integers $b_{i}$ satisfy $0 \leq b_{i} \leq \frac{a_{i}}{k}$ for each $i$. Therefore, there are exactly $\prod_{i=1}^{N}\left(1+\left[\frac{a_{i}}{k}\right]\right)$ such numbers $m$. The value of $P_{k}(n)$ is determined by subtracting the number of those $m$ for which $m^{k+1} \mid n$ from the number of $m$ with the property $m^{k} \mid n$, so

$$
P_{k}(n)=\prod_{i=1}^{N}\left(1+\left[\frac{a_{i}}{k}\right]\right)-\prod_{i=1}^{N}\left(1+\left[\frac{a_{i}}{k+1}\right]\right)
$$

Now we determine the factorization of the number $100 !=2^{a_{1}} 3^{a_{2}} 5^{a_{3}} \ldots$ According to the formula (see, e.g., ŠMM vol. "On Large Numbers", p. 37), the prime $p$ has an exponent in the factorization of $n!$ equal to

$$
\left[\frac{n}{p}\right]+\left[\frac{n}{p^{2}}\right]+\left[\frac{n}{p^{3}}\right]+\left[\frac{n}{p^{4}}\right]+\ldots
$$

(only a finite number of summands are non-zero). Thus, we can establish the prime factorization $100 !=2^{97} 3^{48} 5^{24} 7^{16} 11^{9} 13^{7} 17^{5} \ldots$, where the dots indicate further primes whose exponents are less than 7 and thus do not affect the value of

$$
\begin{aligned}
& P_{7}(100 !)= \\
& =\left(1+\left[\frac{97}{7}\right]\right)\left(1+\left[\frac{48}{7}\right]\right)\left(1+\left[\frac{24}{7}\right]\right)\left(1+\left[\frac{16}{7}\right]\right)\left(1+\left[\frac{9}{7}\right]\right)\left(1+\left[\frac{7}{7}\right]\right)- \\
& -\left(1+\left[\frac{97}{8}\right]\right)\left(1+\left[\frac{48}{8}\right]\right)\left(1+\left[\frac{24}{8}\right]\right)\left(1+\left[\frac{16}{8}\right]\right)\left(1+\left[\frac{9}{8}\right]\right)= \\
& =14 \cdot 7 \cdot 4 \cdot 3 \cdot 2 \cdot 2-13 \cdot 7 \cdot 4 \cdot 3 \cdot 2=4704-2184=2520 .
\end{aligned}
$$

Guiding Problems. (N1) Determine a) the multiplicity of each divisor of the number 144; b) all double divisors of the number $10^{15}$. (N2) Derive formulas (1) and (2).