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

detailed solution

Correct option is C

Note that 7r(r∈N) ends in 7, 9, 3 or 1 (corresponding to  r = 1, 2, 3 and 4 respectively.)Thus 7m+7n cannot end in 5 for any values of m, n ∈ N.In other words, for 7m+7n to be divisible by 5, it should end in 0.For 7m+7n to end in 0, the forms of m and n should be as follows:Thus, for a given value of m there are just 25 values of n forwhich 7m+7n ends in 0. [For instance, if m=4r then n =2, 6, 10, …, 98]∴ there are 100 × 25 = 2500 ordered pairs (m, n) forwhich 7m+7n is divisible by 5.