Question

the hcf of 408 and1032 is expressible in tje form of 1032m-2040 .find the value of m

Answer 1

first find out the hcf
1032=408×2 + 216
408=216×1 + 192
216=192×1 + 24
192=24×8 + 0
so, now since the remainder has come zero 
so the hcf will be 24

now if we put hcf=1032-408×5
1032m-408×5=24
1032m-2040=24
1032m=24+2040⇒2064
m=2064÷1032⇒2
2 is the value of m.

Answer 2

the hcf of 408 and 1032 is 
1032=408×2+216
408=216×1+192
216=192×1+24
192=24×8+0
hence the remainder is 0.
so required hcf is 24

now to express it in given form,
1032m-2040=24
1032m=24+2040
1032m=2064
m=2064/1032
m=2