Question

find the HCF of 52 and 17 and express it in the form 52x + 117y.​

Answer 1

Answer:

X=-2, y=1

Step-by-step explanation:

a=bq+r,0<=r<b

117=52*2+13

52=13*4+0

13=(117*1)-(52*2)

13= - (52*2)+(117*1)

13=52x+117y

X= - 2 and y=1