If the HCF of 65 and 117 is

$\Rightarrow \quad 65=52 \times 1+13$

$\Rightarrow \quad 52=13 \times 4+0$

$\therefore \quad \operatorname{HCF}(65,117)=13$ $\ldots$ (i)

Also, given that, $\operatorname{HCF}(65,117)=65 m-117$ .....(ii)

From Eqs. (i) and (ii),

$65 m-117=13$

$\Rightarrow \quad 65 m=130$

$\Rightarrow \quad m=2$