2. Katya decided to calculate the sum of the cubes of all natural divisors of some natural number, and she got the result $M A T H$. But then she discovered that she had forgotten one of the divisors. Adding its cube, she got the correct result - MASS. Find the smallest possible value of the number $M A T H$. (MATH and $M A S S$ are four-digit numbers, where each digit is replaced by a letter, with the same digits replaced by the same letters, and different digits by different letters.)
Solution. Answer: 2017. The original natural number is $12 ; 12^{3}+6^{3}+4^{3}+2^{3}+1^{3}=$ 2017; if you add $3^{3}$, you get 2044.

We will prove that there are no smaller suitable numbers.

1) For any number less than 10, the sum of the cubes of all divisors, as is easily verified, is less than a thousand.
2) $10^{3}+5^{3}+2^{3}+1^{3}=1134$ and $11^{3}+1^{3}=1332$ cannot equal the number $M A S S$ (the last two digits are not equal).
3) For the number $12 M A S S=2044$; if you subtract a cube less than 27 from it, the result will be greater than 2017, and if a larger one, the result will already be less than 2000.
4) For $n \geqslant 13$ we get: $M A S S \geqslant n^{3} \geqslant 2197$, and therefore, $M A T H \geqslant 2100$.