The number of ways of selecting two numbers a and b, a ∈{2,4,6, … ., 100} and b ∈{1,3,5, … . ., 99} such that 2 is the remainder when a+b is divided by 23 is :

Solution

Maths sir · 3 steps

Even plus odd is always odd, so keep only the odd sums that leave remainder 2, then count pairs for each.

1. Possible sums. Here a is even and b is odd, so a+b is odd and lies between 3 and 199. Numbers that leave remainder 2 on division by 23 are 2, 25, 48, 71, 94, 117, 140, 163, 186, …. The odd ones in range are 25, 71, 117, 163.
2. Count each sum. For a fixed sum s, every even a fixes b = s - a, which is automatically odd. You only need 2 ≤ a ≤ 100 and 1 ≤ b ≤ 99, which means a ≥ s - 99 and a ≤ s - 1.

   Sum a+b	Allowed even a	Number of pairs
   25	2, 4, …, 24	12
   71	2, 4, …, 70	35
   117	18, 20, …, 100	42
   163	64, 66, …, 100	19

3. So. Total = 12 + 35 + 42 + 19 = 108, which is option (c).

The idea to keep. For conditions on a sum, list the allowed sums first and use parity to throw some out. Then for each sum, count one variable and check that the other stays inside its range.

Answer from the NTA final answer key. Solution by PrepWiser.