153. The product of three prime numbers. A certain number is the product of three prime factors, the sum of the squares of which is 2331. There are 7560 numbers (including 1) less than this number and coprime with it. The sum of all divisors of this number (including 1 and the number itself) is 10560. Find this number.
153. Let $N=p q r$. Then $p^{2}+q^{2}+r^{2}=2331$; hence, each of these prime numbers is less than $(2331)^{\frac{1}{2}}<49$, and all prime numbers are odd.

The sum of all divisors of the number $N$ is $(1+p) \cdot(1+q) \cdot(1+r)=$ $=10560=11 \cdot 960$. The only multiple of 11 not exceeding 49 and greater than some prime number by 1 is 44, so[^30]$r=43$. Therefore, $p^{2}+q^{2}=482$ and each of these two numbers is less than $(482)^{\frac{1}{2}}<22$. Next, note that the square of an odd number can end only in 1, 5, or 9, so both $p^{2}$ and $q^{2}$ end in 1. Consequently, $p=11, q=19$ and $N=11 \cdot 19 \cdot 43=8987^{\star}$.