Question

If HCF of 152 and 272 is expressible in the form of 272×8+152×x find the value of x

Answer 1

If the HCF of 408 and 1032 can be written as
By Euclid 's division algorithm,
1032 = 408×2 + 216
408 = 216×1 + 192
216 = 192×1 + 24
192 = 24×8 + 0
Since the remainder becomes 0 here, so HCF of 408 and 1032 is 24
Now,
1032x - 408*5 = HCF of these numbers
=> 1032x - 2040 = 24
=> 1032x = 24+2040
=> x = 2064/1032
=> x = 2
So value of x is 2.

Answer 2

a=bq+r
272=152x1+120
152=120x1+32
120=32x3+24
32=24x1+12
24=12x2+0
272*8+152x=8
152x=8-272*8=-216.8
x=-216.8/152=1.426315789473684=approx 1