17. How many different pairs of integers $A, B$ exist for which the least common multiple is 59400000?
17. The prime factorization of the number 59400000 is

$$
59400000=2^{6} \cdot 3^{3} \cdot 5^{5} \cdot 11
$$

Therefore, if the least common multiple of two numbers $A$ and $B$ is 59400000, then it must be

$$
A=2^{x_{1}} \cdot 3^{\beta_{1}} \cdot 5 \gamma_{1} \cdot 11^{\delta_{1}}, \quad B=2^{\alpha_{2}} \cdot 3^{\beta_{2}} \cdot 5 \tau_{2} \cdot 11^{\delta_{3}}
$$

In this case, it must be:

$\alpha_{1} \leqslant 6, \alpha_{2} \leqslant 6, \max \left(\alpha_{1}, \alpha_{2}\right)=6 ; \beta_{1} \leqslant 3, \beta_{2} \leqslant 3, \max \left(\beta_{1}, \beta_{2}\right)=3 ;$ $\gamma_{1} \leqslant 5, \gamma_{2} \leqslant 5, \max \left(\gamma_{1}, \gamma_{2}\right)=5 ; \hat{c}_{1} \leqslant 1, \hat{c}_{2} \leqslant 1, \max \left(\hat{c}_{1}, \delta_{2}\right)=1$, where $\max (a, b)$ denotes the maximum of two numbers $a$ and $b$.

These conditions are satisfied by 13 possible pairs of values $\alpha_{1}, \alpha_{2}\left(\alpha_{1}=6, \alpha_{2}=0,1,2,3,4,5 ; \alpha_{1}=0,1,2,3,4,5\right.$, $\left.\alpha_{2}=6 ; \alpha_{1}=\alpha_{2}=6\right) ; 7$ possible pairs of values $\beta_{1}, \beta_{2}\left(\beta_{1}=3\right.$, $\left.\beta_{2}=0,1,2 ; \beta_{1}=0,1,2, \beta_{2}=3 ; \beta_{1}=\beta_{2}=3\right) ; 11$ possible

1) Another example of a multiplicative function of a number $N$ is the Euler function $\varphi(N)$ - the number of numbers less than $N$ and coprime with $N$ (see note to the solution of problem 12).
pairs of values $\gamma_{1}, \gamma_{2}\left(\gamma_{1}=5, \gamma_{2}=0,1,2,3,4 ; \gamma_{1}=0\right.$, $1,2,3,4, \gamma_{2}=5 ; \gamma_{1}=\gamma_{2}=5$ ) and 3 possible pairs of values $\delta_{1}, \delta_{2}\left(\hat{\delta}_{1}=1, \delta_{2}=0 ; \delta_{1}=0, \delta_{2}=1 ; \delta_{1}=\delta_{9}=1\right)$. By combining all these values, we get a total of

$$
13 \cdot 7 \cdot 11 \cdot 3=3003
$$

pairs of numbers $A, B$.

However, in our count, two pairs of numbers that differ only in order are considered different. Therefore, each pair of numbers, having the same least common multiple 59400000, except for the pair $A=B=59400000$, is counted twice. Therefore, the number of essentially different pairs of numbers $A, B$ will be

$$
\frac{3003-1}{2}+1=1502
$$

Note. In the same way, it can be shown that if the prime factorization of a number $N$ is

$$
N=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}
$$

then the number $m$ of all possible pairs of integers, the least common multiple of which is $N$, can be calculated by the formula

$$
m=\frac{\left(2 x_{1}+1\right)\left(2 x_{2}+1\right) \ldots\left(2 x_{k}+1\right)-1}{2}+1
$$