The number of ordered pairs (m, n), where m, n ∈{1, 2, 3, …… ,50}, such that 6^m + 9^n is a multiple of 5 is –
(A) 1250 (B) 2500 (C) 625 (D) 500

• m,n(1,2,3...50) 9^m+6^n is multiple of 5
  9^m always unit digit ends with 1 or 9 ...1
  6^n always unit digit ends with 6...2
  5 multiples ends 0 or 5
  from 1 and 2 -9^m+6^n should end with 15
  9^m+6^n should with 15
  possible values of m=1,3,....49 =25
  possible values of n=1,2,3...50
  combinations 25 *50=1250 (b)
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