Question

Prove that if either of $2a+3b$ and $9a+5b$ is divisible by 17 ,so is the other ,$a,b\in N$

Answer 1

Let $2a+3b=17k$

Then $a=\frac{(17k-3b)}{2}$ ----- (i)

Consider,

$9a+5b$

$=\frac{(153k-27b)}{2}+5b$ ---from (i)

$=\frac{(153k-17b)}{2}$

$=\frac{17(9k-b)}{2}$

which is divisible by 17

and vice versa