Question

Express HCF of F 65 and 117 in the form of 65m +117n

Answer 1

117 = 65 x 1 + 52 ; 65 = 52 x 1 + 13 ; 52 = 4 x 13 + 0
In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers
The H.C.F. of 65 and 117 is 13
Now, 13 = 65 – 52 x 1 ⇒ 52 = 117 – 65 x 1 ⇒ 13 = 65 – (117 – 65 x 1) x 1 ⇒13 = 65 x 2 – 117
⇒13 = 65 x 2 + 117 x (–1)
the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1