If the integer s m and n are chosen at random between 1 and 100, then the probability that a number of the form 7m+7n is divisible by 5 equals

We know that 71=7,72=49,73=343,74=2401,75=16807. Therefore 7k(where k∈Z) results in a number whose unit's digit is 7 or 9 or 3 or 1. Now 7m+7n  will be divisible by 5 if unit's place digit in the resulting number is 5 or 0. Clearly, it can never be 5. But it can be 0 if we consider values of m and n such that the sum of unit's place digits become 0. And this can be done by choosingm=1,5,9,.... 97n=3,7,11,...., 99 (25 options each) [7+3=10]orm=2, 6, 10, ....,   98n=4,  8, 12,...., 100 (25 options each) [9+3=10]Therefore, the total number of selections of m, n such that 7m+7n is  divisible by 5 is 25×25+25×25100×100=14