## Task 1 - 320831

If $a, b, c$ are the hundreds, tens, and units digits of a three-digit natural number, then this number is briefly denoted by $\overline{a b c}$. Similarly, a two-digit number with tens and units digits $b$ and $c$ is denoted by $\overline{b c}$.

Determine all $a, b, c$ for which $\overline{a b c}$ is a three-digit number and $\overline{b c}$ is a two-digit number, such that the equation $\overline{a b c}=(\overline{b c})^{b}$ holds!
I. If $a, b, c$ are digits of the required type, then:

It cannot be $b=0$, because then $\overline{b c}$ would not be a two-digit number. It cannot be $b=1$, because then $(\overline{b c})^{b}=\overline{b c}$, which could not be equal to the three-digit number $\overline{a b c}$.

It cannot be $b \geq 3$, because then $(\overline{b c})^{b} \geq 30^{3}$ would be at least a five-digit number, and thus could not be equal to the three-digit number $\overline{a b c}$. Therefore, $b=2$ must hold, and thus $\overline{a b c}=(\overline{b c})^{2}$ must hold.

Therefore, $c$ must be the units digit of the number $c^{2}$. Since $2^{2}=4, 3^{2}=9, 4^{2}=16, 7^{2}=49, 8^{2}=64, 9^{2}=81$, this cannot be true for $c=2 ; 3 ; 4 ; 7 ; 8 ; 9$, and thus can only be true for $c=0 ; 1 ; 5 ; 6$.

Furthermore, from the tens digits of $20^{2}=400, 21^{2}=441, 26^{2}=676$ it is clear:

The number $(\overline{b c})^{2}$ with $b=2$ does not have the required tens digit 2 for $c=0 ; 1 ; 6$ as required by $\overline{a b c}=(\overline{b c})^{2}$; it can only have this (among the possibilities $c=0 ; 1 ; 5 ; 6$) for $c=5$.

This finally leads to $25^{2}=625$, giving $a=6$.

II. From $a=6, b=2, c=5$, the required equation $\overline{a b c}=(\overline{b c})^{b}$ is satisfied because $625=25^{2}$.

With I. and II., it is shown: The digits $a=6, b=2, c=5$ are exactly the digits of the required type.