Question

HCF pf 1624 and 1143​

Answer 1

Answer:

So the HCF of 1624 and 1143 is

Step-by-step explanation:

a=b×q+r (where a is 1624 and b is 1143 and q qoutient and r is remainder )

then

1624=1143×1+481 ( still the r is not zero )

so now a =1143 and b =481

1143=481×2+181 ( still the r is not zero )

so now we a = 481 and b = 181

481= 181 ×2 +119( still the r is not zero )

so now we a= 181 and b=119

181=119×1+62( still the r is not zero )

so now we a =119 and b=62

119=62 ×1+57( still the r is not zero )

so now we a=62 and b= 57

62=57×1+5( still the r is not zero )

so now we a=57and b=5

57=5×10+7 ( still the r is not zero )