If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form $7^m +7^n$ is divisible by 5, is

$\frac{1}{4}$

We have,

Number of ways of selecting m and n = 100 × 100

We know that $7^k , k ∈ N $ has 1, 3, 9, 7, at the units place for

$k = 4λ, 4λ -1, 4λ -2, 4λ - 3.$

Therefore, $7^m + 7^n $ can have, 0, 2, 4, 6 and 8 at units place. But, $7^m + 7^n $ will be divisible by 5 if it has 0 at units place.

The digit at units place in $7^m + 7^n$ will be 0 if m and n have the following forms:

Clearly, for each value of m there are 25  values of n for which the digit at units place in $7^m + 7^n $ is 0.

Number of ways of selecting m and n so that $7^m + 7^n$ is divisible by 5 is

$ 25×25 + 25× 25 + 25 × 25 + 25 × 25 = 2500$

Hence, required probability $=\frac{2500}{100×100}=\frac{1}{4}$