PROBLEM 3. Determine how many numbers of the form $\overline{a b c}$, with distinct digits, satisfy the relation:

$$
\overline{a b a}-\overline{a a a}+17(b-a)=c^{3}
$$
## PROBLEM 3.

(1p) We observe that $b>a>0$

(2p) The relation from the statement is equivalent to: $10 b-10 a+17(b-a)=c^{3}$

(1p) From which we have: $27(b-a)=c^{3} \Rightarrow b-a$ is a perfect cube

(2p) But, $b-a<9 \Rightarrow b-a \in\{1,8\} \Rightarrow c^{3} \in\left\{3^{3}, 6^{3}\right\}$

(1p) Thus, $\overline{a b c} \in\{123,453,563,673,783,893,196\}$, meaning there are 7 numbers.