Question

Express the $H.C.F$ of $48$ and $18$ as a linear combination.

Answer 1

$a=bq+r$, where $0\le r<b$

$48=18\times 2+12$

$18=12\times 1+6$

$12=6\times 2+0$

$\therefore HCF(18,48)=6$

Now,

$6=18-12\times 1$

$6=18-(48-18\times 2)$

$6=18-48\times 1+18\times 2$

$6=18\times 3-48\times 1$

$6=18\times 3+48\times (-1)$

$\Rightarrow 6=18x+48y$

Whereas,

$x=3,y=-1$

Therefore,

$6=18\times 3+48\times (-1)$

$6=18\times 3+48\times (-1)+18\times 48-18\times 48$

$6=18(3+48)+48(-1-18)$

$6=18\times 51+48\times (-19)$

$\Rightarrow 6=18x+48y$

Whereas,

$x=51,y-19$

Thus $x$ and $y$ are not unique.