MathJax TeX Test Page If the integers $m$ and $n$ are chosen at random from $1$ to $100$, then find the probability that the number of the form $7^n+7^m$ is divisible by $5$. $$.$$ Number of numbers of the form $7^n+7^m$ is 100×100= $100^2$ , since m,n ∈{1 ,…………,100}. Now $7^1=7,7^2=49,7^3=343$ and $7^4=2401$ , the digits at the unit places are 7,9,3 and 1. $$.$$ For n ≥5 the digit at the unit place for the number $7^n$ will be one of among the numbers 1,3,7 and 9 $$.$$ A number is divisible by 5 iif the digit at the unit’s place is either 0 or 5. Therefore a number of the form $7^n+7^m$ will be divisible by 5 iff the digits at the unit place of them add up to 10. The possible cases being {1,9} and {3,7} in any order. $$.$$ Since there are 4 distinct digits at the unit place for the number $7^n$ and they are repeated modulo 4, we have 100/4=25 distinct number of the form $7^n$ having the digit 1 or 3 or 7 or 9 at the unit place. $$.$$ When $7^n$ has the digit 1 at the unit place, $7^m$ need to have the digit 9 at the unit place giving 25 such choices. In total 2 x 25 x 25 choices ( n and m can be interchanged!) $$.$$ Similarly for the pair {3,7} we have $2 \times 25 \times 25$ choices, in total giving $4 \times 25 \times 25$ choices!! $$.$$ Required probability $$= {4 \times 25 \times 25 \over 100 \times 100} = {1 \over 4} .$$