A number is the product of four prime numbers. Which is this number, if we know that the sum of the squares of the four prime numbers is 476?
Let the four prime numbers be $p_{1}, p_{2}, p_{3}$, and $p_{4}$. According to the condition,

$$
p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}=476
$$

A square number, when divided by 3, leaves a remainder of 0 or 1. The remainder of 476 when divided by 3 is 2, so exactly two terms of the sum on the left side must be divisible by 3, and since they are prime numbers, $p_{1}=3$ and $p_{2}=3$, from which

$$
p_{3}^{2}+p_{4}^{2}=458
$$

Let $p_{3} \leq p_{4}$. Then $p_{4}^{2} \geq 229$, so $p_{4}>15$, and on the other hand, $p_{4}^{2}<458$, so $p_{4} \leq 21$. Since $p_{4}$ is a prime number, its value can only be 17 or 19. If $p_{4}=17$, then $p_{3}=13$; if $p_{4}=19$, then $p_{3}$ would not be an integer. Therefore, $p_{1}=3, p_{2}=3, p_{3}=13$, $p_{4}=17$, and their product is 1989.

Katz Sándor (Bonyhád, Petőfi S. Gimn., I. o. t.)