Let A={1,2,3,4} and R be a relation on the set A × A defined by
R={((a, b),(c, d)): 2 a+3 b=4 c+5 d}. Then the number of elements in R is ____________.

Solution

Maths sir · 4 steps

The left side can never go past 20, so only a handful of (c,d) pairs are even worth checking.

1. Bound the left side. With a, b ∈ {1,2,3,4}, the smallest value of 2a+3b is 2+3=5 and the largest is 8+12=20. So any match needs 4c+5d ≤ 20.
2. Shortlist (c, d). For d=1, c can be 1, 2 or 3 (c=4 gives 21). For d=2, c can be 1 or 2. For d=3, only c=1 works. For d=4, even c=1 gives 24. That leaves six pairs.
3. Match each value. For each target, solve 2a+3b = target with a, b from 1 to 4. Since 2a is even, 3b must have the same parity as the target, so b is odd for an odd target and even for an even target.

   (c, d)	4c+5d	(a, b) with 2a+3b equal to it	Count
   (1, 1)	9	(3, 1)	1
   (2, 1)	13	(2, 3)	1
   (3, 1)	17	(4, 3)	1
   (1, 2)	14	(4, 2) and (1, 4)	2
   (2, 2)	18	(3, 4)	1
   (1, 3)	19	none: b=1 needs a=8, b=3 needs a=5	0

4. So. Each match is one ordered pair ((a,b),(c,d)) in R. Adding the counts gives 1+1+1+2+1+0 = 6.

The idea to keep. When a relation is defined by an equation on a small finite set, find the range of the smaller side first. That cuts the search down, and a parity check finishes each case fast.

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