probability of divisibility by $5$ Let $m,n$ be $2$ numbers between $1-100$ . what is the probability that if we select any two random numbers then $5|(7^m+7^n)$ . My attempt last digit should be $5$ or $0$ so $7$ powers follow the pattern $7,9,3,1,7...$ so $m,n$ should be such that if one gives $7$ as last digit other should give $3$ as last digit. Same for $9,1$ . but this will take time and it was asked in a competitive exam with little space for Rough work so i believe there exists a simple logical way. Hope you guys help.

As you would have noticed by now, $7^k\equiv7^{k\bmod4}$.
And as you noted, $5|(7^m+7^n)\iff|m-n|\equiv2\pmod4$.
The probability for that in the case of $m,n\in\mathbb{N}$ is obviously $\frac14$.
Since $4|100$, the probability in the case of $m,n\in[1,100]$ is also $\frac14$.
