# 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 equals.

$\begin {array} {1 1} (A)\;\large\frac{1}{4} & \quad (B)\;\large\frac{1}{7} \\ (C)\;\large\frac{1}{8} & \quad (D)\;\large\frac{1}{49} \end {array}$

Total number of cases = $100 \times 100$
Now $7^m + 7^n$ is divisible by 5 if one of the term has to end with 9 and other with 1.
$7^x$ cannot be divisible by 9.
$\therefore$ m can be 2,6,10,14....98 (25 values ) and n can be 4, 8, 12....100 ( 25 values ).
Since m and n can interchange
$\therefore$ Required probability = $\large\frac{ 2 \times 25 \times 25}{100 \times 100}$
= $\large\frac{1}{8}$
Ans : (C)