## Task 1 - 220931

Determine all three-digit numbers $z$ (written in the decimal system) that satisfy the equation $z=(a+b)^{c}$, where $a, b$, and $c$ are the digits of $z$ in any order.
It is $c>0$, since for all $a, b$ it holds that $(a+b)^{0}=1$, which is not a three-digit number. It is also $c>1$ because otherwise $z=(a+b)^{c}$ would not be a three-digit number, given $a+b \leq 18$.

If $c=2$, then due to $10^{2}=100 \leq z=(a+b)^{2}$, it follows that $10 \leq a+b \leq 18$. The resulting square numbers are $10^{2}=100, 11^{2}=121, 12^{2}=144, 13^{2}=169, 14^{2}=196, 15^{2}=225, 16^{2}=256, 17^{2}=289$, and $18^{2}=324$. From these, $100, 144, 169$, and $196$ are excluded because they do not contain the digit $c=2$. For the remaining numbers, it must be checked whether the other two digits, besides the two (or one of the twos), can be chosen as $a$ and $b$. This is not the case for $121=11^{2}$ (since $1+1 \neq 2$), $256=16^{2}$ (since $5+6 \neq 16$), or $324=18^{2}$ (since $3+4 \neq 18$), but it is the case for $17^{2}=289$ (since $8+9=17$). Therefore, $z=289$ is the only solution in this case.

If $c=3$, then due to $4^{3}=64$, $z$ must be at least $5^{3}=125$. Since $z$ must be a three-digit number, $c \leq 9$. Therefore, the cases $2^{7}=128, 2^{8}=256$, and $2^{9}=512$ need to be considered, but none of these provide a solution because none of the powers contain the corresponding $c$ as a digit.

Thus, there are exactly two solutions, namely $z_{1}=289=(8+9)^{2}$ and $z_{2}=343=(3+4)^{3}$.