The correct option is A 14
71=7, 72=49, 73=343, 74=2401,⋯
Therefore, for 7r, r ϵ N the number ends with 7, 9, 3, 1, 7, . . .
∴ 7m+7n will be divisible by 5 if it ends in 0 or 5.
The last digit of m and n belongs to the set {1,3,7,9}
Hence there are 4×4=16 combinations of digits which determines the last digit of the sum.
The favourable cases among them are { (1,9),(9,1),(3,7),(7,3) }
Hence, the required probability is
P=416=14