Divisible by 5 and 39 <=> Divisible by 5,3 and 13.

Because it is divisible by 5 => y = 0 or 5.

If y = 0 then replace: 2 + 3 + x + 8 + 0 \(⋮3\) <=> 13 + x \(⋮3\)

We have x = 2;5;8 satisfy but only x = \(\varnothing\) satisfy divisible by 13.

If y = 5 then replace: 2 + 3 + x + 8 + 5 \(⋮3\) <=> 18 + x \(⋮3\)

We have x = 0;3;6;9 satisfy but x = 9 satisfy divisible by 13

So x = 9 and y = 5 to satisfy divisible by 5 and 39 \(\left(23985⋮5and39\right)\)

\(\overline{23x8y}⋮5\Rightarrow y=0;5\)

\(\overline{23x8y}⋮39\), so it's divisible by 3 and 13

If y = 0, \(\overline{23x80}⋮3\) or \(\left(13+x\right)⋮3\). So, \(x=2;5;8\)

We see that \(\overline{23x80}⋮̸\)\(13\) with x = 2 ; 5 ; 8

If y = 5, \(\overline{23x85}⋮3\) or \(\left(18+x\right)⋮3\). So, x = 0 ; 3 ; 6 ; 9

We see that \(\overline{23x85}⋮13\) only when x = 9

Hence, x = 9 ; y = 5