Question

If \[d\] is the HCF of 56 and 72, find $x,y$ satisfying $d=56x+72y$. Also, show that $x$ and $y$ are not unique.

Answer 1

Hint: Try to use Euclid’s division lemma for finding H.C.F of given numbers. And hence form equation $d=56x+72y$. Now, try to get values of $x$ and $y$ with the equation of Euclid’s division lemma and show that x and y are not unique.

Complete step-by-step answer:

First of all, let us find the H.C.F of 56 and 72 which is represented by \[d\] in the problem.
We know the Euclid’s division lemma as
\[a=bq+r\ldots \ldots (1)\]
where $r$= Remainder, $q$= Quotient, $b$= Divisor, $a$= Dividend
So, we can find H.C.F of 56 and 72 by using Euclid’s division lemma as
$72=56\times 1+16\ldots \ldots (2)$
Now, we need to divide ‘56’ by ‘16’, and we get equation by Euclid’s division lemma as
$56=16\times 3+8\ldots \ldots (3)$
Now, we need to divide ‘16’ by ‘8’, hence, we get
$16=8\times 2+0\ldots \ldots (4)$
Now, the remainder becomes zero. So, H.C.F of 56 and 72 is 8. Hence, the value of \[d\] is 8.
Now, we can form an equation of type
$d=56x+72y$
$56x+72y=8\ldots \ldots (5)$
Now, from equation (4), we get
$8\times 2=16$
Putting value of ‘8’ from equation (3) in above equation, we get
$\left( 56-16\times 3 \right)\times 2=16$
$\left( 56-16\times 3 \right)=8$.
Now, put value of ‘16’ from equation (2), we get
$\left( 56-\left( 72-56 \right)\times 3 \right)=8$
$56\times 4-72\times 3=8$
$\left( 56 \right)\left( 4 \right)+\left( 72 \right)\left( -3 \right)=8\ldots \ldots (6)$
Now, compare equation (6) and (5), we get
$x=\left( 4 \right),y=\left( -3 \right)$
Now, let us add $\left( 56\times 72 \right)$ and subtract the same as well in equation (6), we get
$\left( 56 \right)\left( 4 \right)+\left( 72 \right)\left( -3 \right)+\left( 56\times 72 \right)-\left( 56\times 72 \right)=8$
Above equation can be reformed as
\[56\times 4+56\times 72+72\times \left( -3 \right)+72\times \left( -56 \right)=8\]
$56\times \left( 76 \right)+72\times \left( -59 \right)=8$
Now, comparing the above equation with equation (5), we get
$x=76,y=-59$
Hence, there are no fixed values of ‘x’ and ‘y’. We can get infinite pairs of $x$ and $y$ as we have only one equation $d=56x+72y$ with two variables $x$ and $y$.
Therefore, it is proved that $x$ and $y$ are not unique.

Note: Another approach for the equation would be that we can find H.C.F by prime factorization as well. So, 56 and 72 can be written as
$\begin{align}
  & 56=7\times 2\times 2\times 2 \\
 & 72=3\times 3\times 2\times 2\times 2 \\
\end{align}$
H.C.F = common multiple of both = $2\times 2\times 2=8$.
Now, get equations as
$56x+72y=8$
Now, put $x=0$ , get $y=\dfrac{8}{72}$ and put $y=0$ , get $x=\dfrac{8}{56}$ which proves that $x$ and $y$ are not unique. There are infinite pairs of $x$ and $y$. One needs to be very clear with the terms in Euclid’s division lemma equation i.e., \[a=bq+r\] . If we place any value incorrectly, we will have wrong results and further answers will be wrong as well. Hence, be careful while finding H.C.F with help of Euclid’s division lemma.