Two integers m and n are chosen at random between 1 and 100 (including 1 and 100) then the probability that a number of the form $7^m+ 7^n$ is divisible by 5 is:

1/4

Let us sec 71, 72, 73, 74, 75, 76, 77, 78 and the remainders they leave when divided by 5

The remainders are 2, 4, 3, 1, 2, 4, 3, 1,……

Now if $m ∈ \{1, 5, 9, .......97\} = S_1$

7^m leaves 2 as remainder

if $m ∈ \{2, 6, 10, .......98\} = S_2$

7^m leaves 4 as remainder

If $m ∈ \{3, 7, 11, .......99\} = S_3$

7^m leaves 3 as remainder

if $m ∈ \{4, 8,12, .......100\} = S_4$

7^m leaves 1 as remainder

Now for $7^m + 7^n$ to be divisible by 5,

$m ∈S_1, n ∈S_3\, or\, m ∈S_2, n ∈S_4$

$m ∈S_3, n ∈S_1\, or\, m ∈S_4, n ∈S_2$

No. of favourable numbers = 25 × 25 + 25 × 25 + 25 × 25 + 25 × 25 = 4 × 25 × 25

Total no. of numbers = 100 × 100

$⇒P=\frac{4×25×25}{100×100}=\frac{1}{4}$