16*. How many different divisors does the number 86400000 have (including 1 and the number 86400000 itself)? Find the sum of all these divisors.
16. The prime factorization of the number 86400000 is

$$
86400000=2^{10} \cdot 3^{3} \cdot 5^{5}
$$

Therefore, all its divisors have the form $2^{x} \cdot 3^{3} \cdot 5^r$, where $\alpha$ is a non-negative integer not exceeding 10, $\beta$ is a non-negative integer not exceeding 3, and $\gamma$ is a non-negative integer not exceeding 5. Thus, there are 11 possibilities for $\alpha$ (namely, $0,1,2,3,4,5,6,7,8,9,10$), four possibilities for $\beta$, and six possibilities for $\gamma$. Since when forming divisors we can combine all possible values of the exponents $\alpha, \beta$, and $\gamma$ (including the number itself and 1), the number of all divisors is

$$
11 \cdot 4 \cdot 6=264
$$

Now let's find the sum of all divisors. For this, we first determine the sum $s_{1}$ of all divisors of our number that have the form $2^{\text {x}}$. Clearly, $s_{1}$ is equal to

$$
2^{0}+2^{1}+2^{2}+\cdots+2^{10}=\frac{2^{11}-1}{2-1}=2^{11}-1=2047
$$

Now note that the sum of the divisors of our number that have the form $2^{2} \cdot 3^{\gamma_{1}} \cdot 5^{\gamma_{1}}$, where $\beta_{1}$ and $\gamma_{1}$ are fixed numbers, and $\alpha$ takes all possible values, is $s_{1} \cdot 3^{\beta_{1}} \cdot 5^{\gamma_{1}}$ (since we can factor out $3^{\beta_{1}} \cdot 5^{\gamma_{1}}$ when summing). Therefore, the sum of all divisors of the number 86400000 is the product of 2047 by the sum of all divisors of the number $3^{3} \cdot 5^{5}$ (since all divisors of 86400000 that have the form $3^{\beta_{1}} \cdot 5^{\gamma_{1}}$ are divisors of the number $3^{3} \cdot 5^{5}$).

Similarly, it can be shown that the sum of all divisors of the number $3^{3} \cdot 5^{5}$ is the product of the sum $S_{2}$ of all divisors of this number that have the form $3^{\beta}$, by the sum $s_{8}$ of all divisors of the number $5^{5}$. But

$$
\begin{gathered}
s_{2}=3^{0}+3^{1}+3^{2}+3^{3}=\frac{3^{4}-1}{3-1}=\frac{81-1}{2}=40 \\
s_{8}=5^{0}+5^{1}+\ldots+5^{5}=\frac{5^{6}-1}{5-1}=\frac{15625-1}{4}=3906
\end{gathered}
$$

Therefore, the sum of all divisors of the number 86400000 is

$$
s_{1} \cdot s_{2} \cdot s_{\mathrm{s}}=2047 \cdot 40 \cdot 3906=319823280
$$

Let's provide another version of the same solution. Let $a$ and $b$ be two coprime numbers. In this case, it is clear that each divisor of the number $a \cdot b$ is the product of some divisor of the number $a$ and some divisor of the number $b$. From this, it is easy to derive that the number $n(N)$ of all divisors of the number $N$ and the sum $s(N)$ of these divisors have the following property: if $a$ and $b$ are two coprime numbers, then

$$
n(a \cdot b)=n(a) \cdot n(b) \text { and } s(a \cdot b)=s(a) \cdot s(b)
$$

(in number theory, $n(N)$ and $s(N)$, based on this property, are called multiplicative functions of the number $N$). Now, considering that the number of divisors of the numbers $2^{10}, 3^{3}$, and $5^{5}$ are 11, 4, and 6 respectively, and the sum of all divisors of these numbers are 2047, 40, and 3906 respectively, we get:

$$
\begin{aligned}
& n(86400000)=11 \cdot 4 \cdot 6=264 \\
& s(86400000)=2047 \cdot 40 \cdot 3906=319823280
\end{aligned}
$$

Remark. In the same way, it can be shown that if the prime factorization of the number $n$ has the form

$$
n=p_{1}^{x_{1}} \cdot p_{2}^{z_{2}} \ldots p_{k}^{z_{k} k}
$$

then the number of its divisors is

$$
\left(a_{1}+1\right)\left(a_{2}+1\right) \ldots\left(a_{k}+1\right)
$$

and the sum of these divisors is

$$
\left(\frac{p_{1}^{z_{1}+1}-1}{p_{1}-1}\right)\left(\frac{p_{2}^{z_{2}+1}-1}{p_{2}-1}\right) \ldots\left(\frac{p_{k}^{\alpha_{k}+1}-1}{p_{k}-1}\right)
$$