The number of ordered pairs $\left(m,n\right),m,n\in \left\{1,2,\mathrm{.....100}\right\}$ such that $7^m+7^n$ is divisible by 5 is:

NOTE: $7^r\left(r\in N\right)$ ends in 7, 9, 3 or 1 (corresponding to r= 1, 2, 3 and 4 respectively)

Thus $7^m+7^n$ cannot end is 5 for any values of m, n$\in N$ . In other words, for $7^m+7^n$ to be divisible by 5, it should end in 0.

For $7^m+7^n$ to end on 0, the forms of m and n should as follows:

Thus for a given value of m there are just 25 values of n for which $7^m+7^n$ ends in 0 [for instance , if m = 4r, then n = 2, 6, 10, ………..98]

So there are $100\times 25=2500$ ordered pairs (m, n) for which $7^m+7^n$ is divisible by 5.