Question

Find the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively.

Answer 1

Let x be the no:

So, x= 28M+8 = 32N+12

28M-32N=4

7M-8N=1

Minimum M and N satisfying above is:

M=7 and N=6

So, smallest number = 28 x 7 + 8

=196 + 8

=204