Question

Let ${}^{\prime }{d}^{\prime }$ be the HCF of $24$ and $36.$ Find two number $a$ and $b$ such that $d=24a+36b$

Answer 1

Let us consider the problem:

$d=HCF$ of $24$ and $36$

To find $a$ and $b$ such that $d=24a+36b$

Applying Euclid's Division Lemma to $24$ and $36$, we get

$36=24\left(1\right)+12-----\left(1\right)$

$24=2\left(12\right)+0$

Thus,the $HCF$ of $24$ and $36$ is $12$.

From(1), we have

Implies that

$12=36-24$

Implies that

$12=24(-1)+36\left(1\right)$

Implies that

$12=24a+36b$

Where $a=-1$ and $b=1$