Question

If the HCF of $79$ and $97$ is expressible in the form $97n-79m$ ,then the value of $m-n=?$

Answer 1

Using Euclid's Division Lemma,

$\Rightarrow 97=79\times 1+18$ or $18=97-79$ ...(i)

$\Rightarrow 79=18\times 4+7$ or $7=79-(18\times 4)$...(ii)

$\Rightarrow 18=7\times 2+4$ or $4=18-(7\times 2)$...(iii)

$\Rightarrow 7=4\times 1+3$ or $3=7-4$...(iv)

$\Rightarrow 4=3\times 1+1$ or $1=4-3$...(v)

$\Rightarrow 3=1\times 3+0$...(vi)

Therefore HCF of $97$ and $79$ is $1$.

Now, using (v)

$\Rightarrow 1=4-3$

$\Rightarrow 1=4-(7-4)$ or $1=4\times 2-7$ [using (iv)]

$\Rightarrow 1=[18-(7\times 2\left)\right]\times 2-7$ [using (iii)]

or

$18\times 2+7\times (-5)$

$18\times 2+(79-(18\times 4\left)\right)\times (-5)$ [using (ii)]

$\Rightarrow 1=18\times 22+79\times (-5)$ [using (i)]

or

$(97-79)\times 22+79\times (-5)$

$\Rightarrow 1=97\times 22+79\times (-27)$

or

$97\times 22-79\times 27$

Thus, $m=27$ and $n=22$ and therefore

$m-n=5$