The number of ordered pairs (m,n),m,n∈{1,2,.....100} such that 7m+7n is divisible by 5 is:

NOTE: 7r(r∈N) ends in 7, 9, 3 or 1 (corresponding to r= 1, 2, 3 and 4 respectively)Thus 7m+7n cannot end is 5 for any values of m, n∈N . In other words, for 7m+7n to be divisible by 5, it should end in 0.For 7m+7n to end on 0, the forms of m and n should as follows:Thus for a given value of m there are just 25 values of n for which 7m+7n ends in 0 [for instance , if m = 4r, then n = 2, 6, 10, ………..98]So there are 100×25=2500 ordered pairs (m, n) for which 7m+7n is divisible by 5.