Question

Find HCF of $81$ and $237$ and express it as a linear combination of $81$ and $237$ i.e., HCF $\left(81,237\right)=81x+237y$for some $x$ and $y.$ Find the value of $x$ and $y$.

Answer 1

Step 1: Applying Euclid's division lemma to find the HCF:

Taking $237$ as $a$and $81$as $b$in the equation $a=bq+r$

$237=81\times 2+75$ $...\left(1\right)$ $\left[\because \mathrm{Dividend}=\mathrm{Divisor}\times \mathrm{Quotient}+\mathrm{Remainder}\right]$

We have $75$ as a remainder, now $81$ will become dividend and $75$will be divisor for Euclid’s division lemma.

$81=75\times 1+6...\left(2\right)$

We have $6$ as a remainder, now $75$ will become dividend and $6$ will be divisor for Euclid’s division lemma.

$75=6\times 12+3...\left(3\right)$

We have $3$ as a remainder, now $6$ will become dividend and $3$ will be divisor for Euclid’s division lemma.

$6=3\times 2+0...\left(4\right)$

Since we got the remainder as zero. We will stop here. The divisor for the last step or remainder of the previous step is the HCF of the numbers $81$ and $237$,

So, HCF is $3$.

Step 2: Finding the values of $\mathit{x}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{y}\mathbf{}\mathbf{:}$

write $3$ in the form of $81x+237y$

$$\begin{gathered} 3=75-6\times 12\left[From\left(3\right)\right] \\ =75-\left(81-75\times 1\right)\times 12\left[Replace6from\left(2\right)\right] \\ =75-\left(81\times 12-75\times 12\right) \\ =75+75\times 12-81\times 12 \\ =75\left(1+12\right)-81\times 12 \\ =75\left(13\right)-81\times 12 \\ =13\left(237-81\times 2\right)-81\times 12\left[Replace75from\left(1\right)\right] \\ =13\times 237-81\times 2\times 13-81\times 12 \\ =237\times 13-81\left(26+12\right) \\ =237\times 13-81\times 38 \\ =81\times \left(-38\right)+237\times \left(13\right) \\ \Rightarrow 81x+237y=81\times (-38)+237\times \left(13\right) \\ \therefore Hence\mathit{x}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{38}and\mathit{y}\mathbf{}\mathbf{=}\mathbf{13} \end{gathered}$$

Final answer: Therefore, the value of $\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{38}\mathrm{and}\mathbf{y}\mathbf{}\mathbf{=}\mathbf{13}$