Question

Find the hcf of 52and117. Express it in the form 52x +117y

Answer 1

Answer:

A = bq + r

117 > 52

let a= 117, b=52

117 = 52 x 2 + 13 ( 52 is divisor )

52 = 13 x 4 + 0 ,the division process stops here, as remainder becomes 0 ,so HCF is 13 ( divisor )

13 can be also expressed as 52( -2 ) + 117 ( 1)