1. ## Real no.

ques-find the smallest no. which when divided by 28 gives remainder 8,and when divided by32 gives remainder 12.

2. Hello, sumit2009!

This can be solved with Modulo Arithmetic,
. . but here is an algebraic solution.

Find the smallest number which when divided by 28 gives remainder 8,
and when divided by 32 gives remainder 12.
Let $N$ = the number.

Then: . $\begin{array}{cccc}N &=& 28a + 9 & {\color{blue}[1]} \\ N &=& 32b + 12 & {\color{blue}[2]} \end{array}\;\text{ for integers }a,b$

Equate [1] and [2]: . $28a + 8 \:=\:32b + 12 \quad\Rightarrow\quad 7a \:=\:8b + 1$

. . Then: . $a \:=\:\frac{8b+1}{7} \quad\Rightarrow\quad a \:=\:b + \frac{b+1}{7}$

Since $a$ is an integer, $b+1$ must be a multiple of 7.

. . The first time this happens is: . $b \:=\:6$

Substitute into [2]: . $N \:=\:32(6) + 12 \quad\Rightarrow\quad\boxed{ N \:=\:204}$