NIMCET 2008 Question Paper with solutions

Last digit of $7^{4\mathit{p}+1}=7$
Last digit of $7^{4\mathit{p}+2}=9$
Last digit of $7^{4\mathit{p}+3}=3$
Last digit of $7^{4\mathit{p}}=1$
If $\mathit{m}$ is $4\mathit{p}+1,$ then $\mathit{n}$ should be $4\mathit{p}+3$.
So that $7^{\mathit{m}}+7^{\mathit{n}}$ is divisible by 5 and vice-versa also.
Similarly, if $\mathit{m}$ is $7^{4\mathit{p}+2}$ then $\mathit{n}$ should be $7^{4\mathit{p}}$ and vice-versa to be divisible by 5
So, number of ordered pairs $=4\times {(}^{25}{\mathit{C}}_1{\times }^{25}{\mathit{C}}_1)=2500$