Find-the-smallest-number-which-leaves-remainders-8-and-12-when-divided-by-28-and-32-respectively-

Question Number 1077 by Vishal last updated on 07/Jun/15

F i n d t h e s m a l l e s t n u m b e r w h i c h l e a v e s r e m a i n d e r s 8 a n d 12

w h e n d i v i d e d b y 28 a n d 32 r e s p e c t i v e l y .

Commented by prakash jain last updated on 07/Jun/15

I assume you mean smallest + ve integer .

Answered by prakash jain last updated on 07/Jun/15

Let smallest positive integer be n

n = 28 x + 8; n = 32 y + 12 x, y > 0

28 x + 8 = 32 y + 12

7 x + 2 = 8 y + 3

7 x = 8 y + 1

8 y \equiv 6 (\mod 7)

y \equiv 6 (\mod 7)

y = 7 t + 6, t = 0, 1 . .

smallest y = 6

x = 7

n = 28 \times 7 + 8 = 196 + 8 = 204

n = 32 \times 6 + 12 = 192 + 12 = 204

smallest + ve integer n = 204