Question

If $d$ is $\mathrm{HCF}$ of $56$ and $72$, then find $x,y$ satisfying the equation $d=56x+72y$. Also show that $x$ and $y$ are not unique.

Answer 1

Step 1: Calculate $d$ by division.

Since $d$ is the $\mathrm{HCF}$ of $56$ and $72$, it divides both the numbers evenly.

$\begin{array}{ccc}2& 56& 72\\ 2& 28& 36\\ 2& 14& 18\\ & 7& 9\end{array}$

Thus, $\mathrm{HCF}$ of $56$ and $72$ is $2\times 2\times 2=8$.

Step 2: Express $8$ in terms of $56$ and $72$.

Write $8$ in terms of $56$.

Using Euclid's division algorithm,

$\begin{array}{rcl}& & \begin{array}{cc}& 56=16\times 3+8\\ \Rightarrow & 8=56-(16\times 3)\end{array}\end{array}$

Similarly, write $16$ in terms of $72$.

$\begin{array}{ccc}\Rightarrow & 8=56-(72-56)\times 3& \left[\because 16=72-56\right]\\ \Rightarrow & 8=56-72\times 3+56\times 3& \left[\because a(b+c)=ab+ac\right]\\ \Rightarrow & 8=56(1+3)+72(-3)& \\ \Rightarrow & 8=56\left(4\right)+72(-3)& \end{array}$

Step 3: Compare with the given equation.

Comparing $d=56x+72y$ and $8=56\left(4\right)+72(-3)$,

$x=4$ and $y=-3$.

So, the value of $x$ is $4$ and the value of $y$ is $-3$.