15. In how many ways can a million be factored into three factors? Factorizations that differ only in the order of the factors are considered the same.
15. Since a million is equal to $2^{6} \cdot 5^{6}$, any of its divisors has the form $2^{\alpha} \cdot 5^{\beta}$, and the factorization of a million into 3 factors has the form

$$
1000000=\left(2^{\alpha_{1}} \cdot 5^{\beta_{2}}\right)\left(2^{\alpha_{2}} \cdot 5^{\beta_{2}}\right)\left(2^{\alpha_{3}} \cdot 5^{\beta_{3}}\right) ;
$$

here $\alpha_{1}, \alpha_{2}, \alpha_{3} ; \beta_{1}, \beta_{2}, \beta_{3}-$ are non-negative integers satisfying the equations

$$
\alpha_{1}+\alpha_{2}+\alpha_{3}=6, \quad \beta_{1}+\beta_{2}+\beta_{3}=6
$$

Let's count how many such systems of numbers $\alpha_{1}, \alpha_{2}, \alpha_{3}$; $\beta_{1}, \beta_{2}, \beta_{3}$ can be.

If $\alpha_{1}=6$, then $\alpha_{2}$ and $\alpha_{3}$ must be equal to 0; thus, in this case, we have a unique system of numbers $\alpha_{1}, \alpha_{2}, \alpha_{3}$.

If $\alpha_{1}=5$, then there are two systems:

$$
\alpha_{1}=5, \alpha_{2}=1, \alpha_{3}=0 \text { and } \alpha_{1}=5, \alpha_{2}=0, \alpha_{3}=1
$$

If $\alpha_{1}=4$, then there are three systems:

$$
\begin{gathered}
\alpha_{1}=4, \alpha_{2}=2, \alpha_{3}=0 ; \alpha_{1}=4, \alpha_{2}=1, \alpha_{3}=1 \\
\alpha_{1}=4, \alpha_{2}=0, \alpha_{3}=2
\end{gathered}
$$

Similarly, it can be shown that if $\alpha_{1}=3$, there are four systems; if $\alpha_{1}=2$, there are five systems; if $\alpha_{1}=1$, there are six systems, and if $\alpha_{1}=0$, there are seven systems. Therefore, the total number of systems of numbers $\alpha_{1}, \alpha_{2}, \alpha_{3}$ satisfying the equation $\alpha_{1}+\alpha_{2}+\alpha_{3}=6$ is

$$
1+2+3+4+5+6+7=28
$$

Similarly, there are 28 different systems of numbers $\beta_{1}, \beta_{2}$, $\beta_{3}$ satisfying the equation $\beta_{1}+\beta_{2}+\beta_{3}=6$. Since in the factorization of the number 1000000 into three factors, we can combine any triplet of numbers $\alpha_{1}, \alpha_{2}, \alpha_{3}$ with any triplet of numbers $\beta_{1}, \beta_{2}, \beta_{8}$, the total number of factorizations is

$$
28 \cdot 28=784
$$

However, in this count, factorizations that differ only by the order of the factors are considered different, i.e., many factorizations are counted multiple times.

Let's determine how many times each factorization is counted. 1) Only one factorization, namely

$$
10^{8}=\left(2^{2} \cdot 5^{2}\right) \cdot\left(2^{2} \cdot 5^{2}\right) \cdot\left(2^{2} \cdot 5^{2}\right)
$$

is counted once.

2) If in the factorization of $10^{6}$ into three factors, two of them are equal (and the third is different), then such a factorization is counted three times: the factor that is different from the others can stand in the product in the first, second, or third position.

Let's count the number of such factorizations. Suppose the factor that appears twice in the factorization has the form $2^{\alpha} \cdot 5^{\beta}$ (i.e., the factorization has the form $10^{6}=\left(2^{\chi} \cdot 5^{\beta}\right) \cdot\left(2^{\alpha} \cdot 5^{\beta}\right)\left(2^{6-2 \alpha} \cdot 5^{6-2 \beta}\right)$; then $\alpha$ can be $0,1,2$ or $3, \beta$ can also be $0,1,2$ or 3. Since any $\alpha$ can be combined with any $\beta$, the total number of possibilities is $4 \cdot 4=16$. We need to exclude one possibility $\alpha=2, \beta=2$, since in this case we get the factorization

$$
10^{6}=\left(2^{2} \cdot 5^{2}\right) \cdot\left(2^{2} \cdot 5^{2}\right) \cdot\left(2^{2} \cdot 5^{2}\right)
$$

highlighted earlier. Thus, 15 factorizations are counted three times.

3) The remaining factorizations are counted six times. Indeed, if all three factors are different from each other, then the following six factorizations differ only by the order of the factors:

$$
\begin{aligned}
& \left(2^{\alpha_{1}} \cdot 5^{\beta_{1}}\right) \cdot\left(2^{\alpha_{3}} \cdot 5^{\beta_{3}}\right) \cdot\left(2^{\gamma_{3}} \cdot 5^{\beta_{3}}\right) ; \quad\left(2^{\alpha_{1}} \cdot 5^{\beta_{1}}\right) \cdot\left(2^{\pi_{3}} \cdot 5^{\beta_{3}}\right) \cdot\left(2^{\alpha_{3}} \cdot 5^{\beta_{3}}\right) ; \\
& \left(2^{\alpha_{2}} \cdot 5^{\beta_{2}}\right) \cdot\left(2^{\alpha_{1}} \cdot 5^{\beta_{1}}\right) \cdot\left(2^{\alpha_{3}} \cdot 5^{\beta_{3}}\right) ; \quad\left(2^{\alpha_{1}} \cdot 5^{\beta_{3}}\right) \cdot\left(2^{\alpha_{3}} \cdot 5^{\beta_{3}}\right) \cdot\left(2^{\alpha_{1}} \cdot 5^{\beta_{1}}\right) ; \\
& \left(2^{\gamma_{3}} \cdot 5^{\beta_{3}}\right) \cdot\left(2^{\alpha_{1}} \cdot 5^{\beta_{2}}\right) \cdot\left(2^{\alpha_{1}} \cdot 5^{\beta_{2}}\right) ; \quad\left(2^{\alpha_{3}} \cdot 5^{\beta_{3}}\right) \cdot\left(2^{\alpha_{2}} \cdot 5^{\beta_{2}}\right) \cdot\left(2^{z_{1}} \cdot 5^{\beta_{1}}\right) .
\end{aligned}
$$

Therefore, the total number of different factorizations of a million into three factors is:

$$
\begin{aligned}
1+15+\frac{784-15 \cdot 3-1 \cdot 1}{6} & =1+15+\frac{738}{6}= \\
& =1+15+123=139
\end{aligned}
$$