## Task 29/81

We are looking for all (proper) three-digit numbers where the sum of the $i$-th powers of the $i$-th digit (counted from left to right) equals the original number.
The sought numbers have the form $n=100 a+10 b+c$ with $a ; b ; c \in \mathbb{N}, 1 \leq a \leq 9, 0 \leq b ; e \leq 9$. According to the conditions of the problem, the equation

$$
a+b^{2}+c^{2}=100 a+10 b+c \quad \text { or } \quad c^{3}-c=(c-1) c(c+1)=99 a+b(10-b)
$$

holds. The term $b(10-b)$ can only take the values $0 ; 9 ; 16 ; 21 ; 24 ; 25$; since the left side of this equation can be represented as the product of three consecutive natural numbers, it, and thus also the right side, is divisible by 3 without a remainder, and the numbers 16 and 25 are excluded.

Since $a \geq 1, b(10-b) \geq 0$, it follows that $c^{3}-c \geq 99, c^{3}>99$, so $c>5$. One can now try the five possibilities for $c$

| $c$ | $c^{3}-c$ | $a$ | $b(10-b)$ | $b$ |
| :---: | :---: | :---: | :---: | :---: |
| 5 | 120 | 1 | 21 | $3 ; 7$ |
| 6 | 210 | 2 | 12 | excluded |
| 7 | 336 | 3 | 39 | excluded |
| 8 | 504 | 5 | 9 | $1 ; 0$ |
| 9 | 720 | 7 | 27 | excluded |

Thus, there are four such numbers: $n_{1}=135, n_{2}=175, n_{3}=518, n_{4}=598$.