Probability that $7^m+7^n$ is divisible by $5$
If $m,n$ are chosen from the first hundred natural numbers with replacement, the probability that $7^m+7^n$ is divisible by $5$ is?
$$7^m+7^n=7^m(1+7^{n-m}), n\ge m$$ The above expression is divisible by $5$ only if $n-m=4k+2$. The max value of $k$ is $24$.
So is the number of possibilities $48$? ($24$ for each $n>m$ and $m>n$)
Solution 1:
The possibilities of ending digits of $7^m$ or $7^n$ : $9,3,1,7$
Favourable cases : $(9,1),(3,7),(1,9),(7,3)$
Probability : $\frac{4}{4\times 4} =\frac{1}{4}$