Question

If d is the HCF of 160 and 24, then find x and y satisfying $d=160x+24y$

• A. 23 and -153
• B. 43 and -153
• C. 33 and 153
• D. -153 and 23

Answer 1

The correct option is A 23 and -153

Let's first find the HCF of 160 and 24
Applying Euclid's division lemma
$160=24\times 6+16$......(i)
Remainder is not zero, applying again
$24=16\times 1+8$......(ii)
Remainder is not zero, applying again
$16=8\times 2+0$......(iii)
Remainder is 0, so HCF is 8
i.e., d = 8
From (i)
$16=160-24\times 6$.......(iv)
From (ii)
$8=24-16\times 1$.......(v)
Now,
$d=8=24-16\times 1$ [using (v)]
$\Rightarrow d=24-(160-24\times 6)\times 1$ [using (iv)]
$\Rightarrow d=24\times 7$
$\Rightarrow d=160\times (-1)+24\times 7$
Comparing with, $d=160x+24y,$ we get,
$x=-1,y=7$
Also,
$160x+24y=160\times (-1)+24\times 7$
$160x+24y=160\times (-1)+24\times 7+160\times 24-160\times 24$
$=160\times 23+24\times (-153)$
Here, x =23, y = -153
Hence, x and y are not unique. In fact we can get many such values of x and y

So, option a is correct.