Three distinct numbers are selected randomly from the set {1,2,3, …, 40}. If the probability, that the selected numbers are in an increasing G.P., is m/n, gcd(m, n)=1, then m+n is equal to __________ .

Solution

Maths sir · 4 steps

The ratio need not be a whole number: write it as p/q and the G.P. becomes kq², kpq, kp².

1. General form. Let the common ratio be r = p/q in lowest terms, with p > q ≥ 1 because the G.P. is increasing. For all three terms to be integers, the first term must be a multiple of q². So the triple is kq², kpq, kp² for a positive integer k, and the condition is kp² ≤ 40.
2. Count by p. For each p, take every q < p that is coprime to p, and allow k = 1 to ⌊ 40/p² ⌋. From p = 7 onwards, p² = 49 > 40, so nothing more is possible.

   p	Allowed q	Values of k	Triples
   2	1	⌊ 40/4 ⌋ = 10	1 × 10 = 10
   3	1, 2	⌊ 40/9 ⌋ = 4	2 × 4 = 8
   4	1, 3	⌊ 40/16 ⌋ = 2	2 × 2 = 4
   5	1, 2, 3, 4	⌊ 40/25 ⌋ = 1	4 × 1 = 4
   6	1, 5	⌊ 40/36 ⌋ = 1	2 × 1 = 2

3. Probability. Favourable cases are 10 + 8 + 4 + 4 + 2 = 28. Each set of three numbers has exactly one increasing order, so the total is C(40, 3) = 9880. The probability is 28/9880 = 7/2470, and gcd(7, 2470) = 1.
4. So. m = 7 and n = 2470, so m + n = 2477.

The common slip: 4949. Counting only whole-number ratios (2, 3, 4, 5, 6) gives 18 triples and a probability of 9/4940, so 4949. It misses fractional ratios such as 3/2, as in 4, 6, 9.

The idea to keep. Integer G.P.s can have fractional ratios like 3/2 (4, 6, 9). Write the ratio as p/q in lowest terms, so the terms are kq², kpq, kp². Then count using the largest term kp².

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