Question

if hcf of 384 and 26 is 2 find m,n such that 2=m×384+n×26 where m and n are integers.

Answer 1

using Euclid's division lemma
$384 = 26 \times 14 + 20 \\ 26 = 20 \times 1 + 6 \\ 20 = 6 \times 3 + 2 \\ 6 = 2 \times 3 + 0$
hence HCF is 2. Now starts from second last ewuation
$2 = 20 - (6 \times 3) \\ 2 = 20 - (26 - 20)3 \\ 2 = 384 - 26 \times 14 - 26 \times 3 + 20 \times 3 \\ 2 = 384 - 26 \times 17 + (384 - 26 \times 14) \times 3 \\ 2 = 384 + 384 \times 3 - 26 \times 17 - 26 \times 42 \\ 2 = (4)384+ ( - 59) \times 26$
so m=4
n= -59

Answer 2

value of M is minus 2 and n is equal to 30