# Kỹ thuật đồng dư trong APMOPS 2004

Question:

A three digit number $\overline{5ab}$ is written 99 times as $A=$5ab5ab5ab…5ab.
The resultant number is a multiple of 91.
What is the three digit number?

Solution:

It is easy to prove that $A=\overline{5ab}(10^{3.98}+10^{3.97}+...+10^3+1).$ Observe that $10^3=1000\equiv -1 (\emph{ mod 91})$.

Thus $10^{3.98}+10^{3.97}+...+10^3+1\equiv 1(\emph{ mod 91})$ $\Rightarrow 91\mid \overline{5ab}.$

Now we can find that $\overline{5ab}=546.$