153*. The number $N$ has the form $p^{\alpha} q^{\beta} r^{\gamma}$, where $p, q, r$ are prime numbers, and $p q - r = 3, p r - q = 9$; furthermore, the numbers $\frac{N}{p}, \frac{N}{q}, \frac{N}{r}$ have respectively 20, 12, and 15 fewer divisors than the number $N$. Find the number $N$.

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153. From the system $\left\{\begin{array}{l}p q-r=3 \\ p r-q=9\end{array}\right.$, we find $(p+1)(r-q)=6$, hence $\left\{\begin{array}{l}p+1=3 \\ r-q=2\end{array}, p=2 ;\left\{\begin{array}{l}2 q-r=3 \\ r-q=2\end{array}, q=5, r=7\right.\right.$.

The combination $p+1=6$ and $r-q=1$ does not yield the required results. Therefore, the desired number has the form $N=$ $=2^{\alpha} \cdot 5^{\beta} \cdot 7^{\gamma}$. The number of all divisors of $N$ is $(\alpha+1)(\beta+1)(\gamma+1)$.

The number of all divisors for $\frac{N}{2}$ is $\alpha(\beta+1)(\gamma+1)$, for $\frac{N}{5}$ is $\beta(\alpha+1)(\gamma+1)$, and for the number $\frac{N}{7}$ is $\gamma(\alpha+1)(\beta+1)$.

According to the problem, we get the system:

$$
\left\{\begin{array}{l}
(\beta+1)(\gamma+1)=20 \\
(\alpha+1)(\gamma+1)=12 \\
(\alpha+1)(\beta+1)=15
\end{array}\right.
$$

Multiplying, we find $(\alpha+1)(\beta+1)(\gamma+1)=60 ; \quad$ therefore, $\alpha=2, \beta=4, \gamma=3$.

Answer. 857500.