Question

Solve in positive integers:
$23x+17y+11z=130$.

Answer 1

$23x+17y+11z=130$

Here, the maximum value of $x=\left[\frac{130}{23}\right]=5$

If $x=5,17y+11z=15$. This will not have any positive integral solutions.

If $x=4,17y+11z=38$. This will not have any positive integral solutions.

If $x=3,17y+11z=61\Rightarrow y=1$ and $z=4$.

If $x=2,17y+11z=84\Rightarrow y=3$ and $z=3$.

If $x=1,17y+11z=107\Rightarrow y=5$ and $z=2$.