If the HCF of 65 and 117 is expressible in the form 65m -117, then the value of m is
(a) 4
(b) 2
(c) 1
(d) 3
(b) By Euclid's division algorithm,
$b=a q+r, 0 \leq r $\Rightarrow \quad 117=65 \times 1+52$ $\Rightarrow \quad 65=52 \times 1+13$ $\Rightarrow \quad 52=13 \times 4+0$ $\therefore \quad \operatorname{HCF}(65,117)=13$ $\ldots$ (i) Also, given that, $\operatorname{HCF}(65,117)=65 m-117$ .....(ii) From Eqs. (i) and (ii), $65 m-117=13$ $\Rightarrow \quad 65 m=130$ $\Rightarrow \quad m=2$
$\Rightarrow \quad 117=65 \times 1+52$
$\Rightarrow \quad 65=52 \times 1+13$
$\Rightarrow \quad 52=13 \times 4+0$
$\therefore \quad \operatorname{HCF}(65,117)=13$ $\ldots$ (i)
Also, given that, $\operatorname{HCF}(65,117)=65 m-117$ .....(ii)
From Eqs. (i) and (ii),
$65 m-117=13$
$\Rightarrow \quad 65 m=130$
$\Rightarrow \quad m=2$
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