a) Let's determine all the divisors of the number 572 based on its prime factorization.

b) How many divisors does the number $572 a^{3} b c$ have, if

-I. $a, b$ and $c$ are distinct prime numbers greater than 20, or
-II. $a=31, b=32, c=33$?
a) $572=2^{2} \cdot 11 \cdot 13$, the only integer divisors can be those in which no other prime number appears in their factorization except 2, 11, and 13, and these can appear at most 2 times, 1 time, and 1 time, respectively. Therefore, regarding the inclusion of the prime number 2 in the divisor, we have 3 choices: we can include it 2 times, 1 time, or not at all, i.e., 0 times. In other words, the exponent of 2 can be 2, 1, or 0. Similarly, the exponents of 11 and 13 can each be 1 or 0. The exponents of 2, 11, and 13 can be chosen independently, so the total number of possibilities is $3 \cdot 2 \cdot 2=12$, which is the product of the numbers obtained by adding 1 to each of the original exponents. If we do not include any of the prime numbers in the divisor, we get 1, and if we include each with the highest exponent, we get 572 itself. The divisors are as follows:
1,
2,
$2^{2}=4$,
11,
$2 \cdot 11=22$,
$2^{2} \cdot 11=44$,

and their multiples by 13

$$
\begin{array}{rrr}
13, & 2 \cdot 13=26, & 2^{2} \cdot 13=52, \\
11 \cdot 13=143, & 2 \cdot 11 \cdot 13=286, & 2^{2} \cdot 11 \cdot 13=572
\end{array}
$$

or in ascending order:

![](https://cdn.mathpix.com/cropped/2024_05_02_16b33210eccd16d6e2d3g-1.jpg?height=59&width=991&top_left_y=784&top_left_x=556)
b) $572 a^{3} b c=2^{2} \cdot 11 \cdot 13 \cdot a^{3} \cdot b \cdot c$. In the first case, $a, b, c$ are all different from 2, 11, 13, and each other, so the number of divisors, similar to the previous reasoning, is

$$
(2+1)(1+1)(1+1)(3+1)(1+1)(1+1)=192
$$

In the second case, 31 is a prime number, $32=2^{5}, 33=3 \cdot 11$, thus $572 a^{3} b c=2^{7} \cdot 3 \cdot 11^{2} \cdot 13 \cdot 31^{3}$, and the number of divisors is:

$$
(7+1)(1+1)(2+1)(1+1)(3+1)=384
$$

Kálmán Thurnherr (Budapest, Radnóti M. gyak. g. I. o. t.)