# Real no.

• Mar 13th 2009, 04:59 AM
sumit2009
Real no.
ques-find the smallest no. which when divided by 28 gives remainder 8,and when divided by32 gives remainder 12.
• Mar 13th 2009, 06:12 AM
Soroban
Hello, sumit2009!

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

Quote:

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

Let $\displaystyle N$ = the number.

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

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

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

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

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

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