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

Note that $7^r(r\in N)$ ends in 7, 9, 3 or 1

( corresponding to r = 1,2,3 & 4 respectively)

Thus, $7^m+7^n$ cannot end in 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 in 0, the forms of m and n should be as follows.mn

1       4r          4s+2

2       4r+1      4s+3

3       4r+2        4s

4       4r+3        4s+1

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]. There are 100x 25 =2500 ordered pairs (m, n) for which $7^m+7^n$ is divisible by 5.