Question

The number of ordered pairs (m, n), m, n Ī {1, 2, … 100} such that 7^m + 7ⁿ is divisible by 5 is –

• A. 1250
• B. 2000
• C. 2500
• D. 5000

Answer 1

Answer: (C)

2500

Answer 2

Note that 7^r (r Ī N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).

Thus, 7^m + 7ⁿ cannot end in 5 for any values of m, n Ī N. In other words, for 7^m + 7ⁿ to be divisible by 5, it should end in 0.

For 7^m + 7ⁿ to end in 0, the forms of m and n should be as follows :

m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7^m + 7ⁿ ends in 0.

[For instance, if m = 4r, then = 2, 6, 10,….. , 98]

\ There are 100 × 25 = 2500 ordered pairs (m, n) for which 7^m + 7ⁿ is divisible by 5.

Hence (3) is correct answer.