Question

Use Euclid's division algorithm to find the HCF of 96 and 60, hence express the HCF in the form of 96x + 60y, where x and y are integers.

✔️ Explaination needed.
✔️ No Spamming.​

Answer 1

Answer:

According to Euclid division lemma, any number a can be written as where, 0 ≤ r < b

HCF of 96 and 60 by using EDL :

so, we can write 96 in the form 60 by using Euclid division lemma

e.g., 96 = 60 × 1 + 36

again, we can write 60 in the form of 36 by using EDL,

e.g., 60 = 36 × 1 + 24

again, we can write 36 in the form of 24 by using EDL,

e.g., 36 = 24 × 1 + 12

again, we can write 24 in the form of 12 by using EDL,

e.g., 24 = 12 × 2 + 0

here, remainder, r = 0 so, HCF of 96 and 60 = 12

now, we have to write HCF of 96 and 60 in the form of 96x + 60y .

12 = 96x + 60y

or, 1 = 8x + 5y

if we choose x = 2 and y = -3

then, 8 × 2 + 5 × -3 = 1

Hence, HCF of 96 and 60 = 96(2) + 60(-3)

is in the form of 96x + 60y where x and y are integers.

Answer 2

According to Euclid's division lemma ,

any two positive or negative integers , say x and y ,or -x or -y there exist unique positive integers or negative integers say q and r satisfying x = yq + r or , -x = (-yq)+(-r) respectively where 0≤r <y .

By Euclid's division lemma , we get

96= 60×1 + 36 .....(1)

60= 36×1 + 24 ....(2)

36= 24×1 + 12.....(3)

24= 12×2 +0

Here we notice that the remainder is zero , and the divisor at this stage is 12.

So , the HCF is 12 .

From (3), we have

12= 36-24

→ 12= 36-(60-36). {From ,(2)}

→12 = (96-60)-{60-(96-60)} {From ,1)}

→ 12 = (96-60)-(60-96+60)

→12= 96 -60-120+ 96

→12 = 96(2)+60(-3)

→12= 96x +60y , where x = 2 and y =-3