Question

if d is the hcf of 32 and 60,find x and y satisfying d= 32x+60y

Answer 1

Answer:

Step-by-step explanation:

4 is the HCF of 32 and 60

Therefore d= 4

60=32×1+28

32=28×1+4

28=4×7+0

Thus,4=32-28×1

4= 32-(60-32×1)×1

4=32-60-32×1

4=32(2)-60(1)

4=32(2)+60(-1)

4=32(x)+60(y)

x=2 y=-1

Answer 2

Answer:

Step-by-step explanation:

HCF of 32 and 60

By Euclid's division lemma

a=bq+r

60=32(1)+28

32=28(1)+4

28=4(7)+0

HCF=4

GIVEN d=4

4=32-28

32-(60-32)

32-60+32

4=32(2)+60(-1)

COMPARE WITH

d=32(x) +60(y)

SOLUTIONS

d=4

x=2

y=-1