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

Since, $m$ and $n$ are selected between 1 and 100 , hence sample space $=100 \times 100$ Also, $\quad 7^1=1,7^2=49,7^3=343,7^4=2401$ $7^5=16807$ etc.
Hence, 1, 3, 7 and 9 will be the last digits in the power of 7 . Hence, for favourable cases. \begin{array}{lllllll}{c}{nm} \\&& \downarrow & & \\1,2 & 1,2 & 1,3 & \ldots & 1,100 & \\2,1 & 2,2 & 2,3 & \ldots & 2,100 \\100,1 & 100,2 & 100,3 & \ldots & 100,100\end{array}
$\therefore$ favourable cases $=25$ for $m=2, \quad n=4,8,12,10 \quad \ldots \quad 100$ $\therefore$ favourable cases $=25$ Similarly for every $m$, favourable $n$ are 25 . $\therefore$ Total favourable cases $=100 \times 25$ Hence, required probability $=\frac{100 \times 25}{100 \times 100}=\frac{1}{4}$