Question

If h is the hcf of 4052 & 12576 & h=4052a+ 12576b then find value of x and y

Answer 1

hcf (4052, 12576) = h

12576 = 3 * 4052 + 420
  4052 =  420 * 9 + 272
   420  =  272 * 1 + 148
   272 =   148 * 1  + 124
   148  =   124 * 1 + 24
   124   =  5 * 24 +  4
    24     =  6 * 4  +   0
So  4 is the  HCF.


4 = 124 - 5 * 24 
    = 124 - 5 * (148 - 124)
    = 6 * 124 - 5 * 148
    = 6 * (272 - 148) - 5 * 148
     = 6 * 272 - 11 * 148
    = 6 * 272 - 11 * (420 - 272)
     = 17 * 272 - 11 * 420
    = 17 * (4052 - 420 * 9) - 11 * 420
     = 17 * 4052  -  164 * 420 
    = 17 * 4052 - 164 * (12576 - 3 * 4052)
    = 509 * 4052 - 164 * 12576

So     a = 509   and  b = -164.