Question

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

Hint:

When selecting numbers in a geometric progression, consider each possible common ratio and check for conditions on \( a \).

Answer 1

Correct Answer: 2477

Determine the probability that three distinct numbers randomly selected from the set \( \{1, 2, 3, ..., 40\} \) form an increasing Geometric Progression (G.P.). Express this probability as a fraction \( \frac{m}{n} \) in simplest form, and subsequently compute \( m + n \).

Key Concepts:

The solution employs the following principles:

1. Combinations: The number of ways to select \(k\) items from a set of \(n\) distinct items, denoted as \( \binom{n}{k} \), is calculated using the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).

2. Geometric Progression (G.P.): Three distinct integers \(a, b, c\) constitute an increasing G.P. if \( b^2 = ac \) and \( a < b < c \). The common ratio, \( r = \frac{b}{a} = \frac{c}{b} \), must exceed 1. As \(a, b, c\) are integers, \(r\) must be a rational number, representable as \( r = \frac{p}{q} \) where \(p\) and \(q\) are coprime integers with \( p > q \ge 1 \). The terms of such a G.P. are \( a, a\frac{p}{q}, a\frac{p^2}{q^2} \). For all terms to be integers, \( q^2 \) must be a divisor of \( a \). Therefore, we can express \( a \) as \( a = k \cdot q^2 \) for some integer \( k \ge 1 \). This results in the triplet form \( (kq^2, kqp, kp^2) \).

Solution Steps:

Step 1: Calculate the total number of possible outcomes.

The total number of ways to select 3 distinct numbers from a set of 40 is given by:

\[ N_{\text{total}} = \binom{40}{3} = \frac{40 \times 39 \times 38}{3 \times 2 \times 1} = 40 \times 13 \times 19 = 9880 \]

Step 2: Determine the number of favorable outcomes by identifying all increasing G.P. triplets.

We seek triplets \( (a, b, c) \) within \( \{1, 2, ..., 40\} \) such that \( a < b < c \) and \( b^2 = ac \). These can be systematically found using the triplet form \( (kq^2, kqp, kp^2) \), subject to the constraint that the largest term, \( kp^2 \), does not exceed 40.

Step 3: Enumerate triplets with an integer common ratio (\( q=1 \)).

These triplets are of the form \( (k, kr, kr^2) \) where \( kr^2 \le 40 \).

For \( r=2 \): \( 4k \le 40 \implies k \le 10 \). (10 triplets)

For \( r=3 \): \( 9k \le 40 \implies k \le 4 \). (4 triplets)

For \( r=4 \): \( 16k \le 40 \implies k \le 2 \). (2 triplets)

For \( r=5 \): \( 25k \le 40 \implies k \le 1 \). (1 triplet)

For \( r=6 \): \( 36k \le 40 \implies k \le 1 \). (1 triplet)

Total for integer ratios: \( 10 + 4 + 2 + 1 + 1 = 18 \).

Step 4: Enumerate triplets with a non-integer rational common ratio (\( q>1 \)).

These triplets are of the form \( (kq^2, kqp, kp^2) \) where \( kp^2 \le 40 \).

For \( r = \frac{3}{2} \): Triplet \( (4k, 6k, 9k) \). \( 9k \le 40 \implies k \le 4 \). (4 triplets)

For \( r = \frac{4}{3} \): Triplet \( (9k, 12k, 16k) \). \( 16k \le 40 \implies k \le 2 \). (2 triplets)

For \( r = \frac{5}{2} \): Triplet \( (4k, 10k, 25k) \). \( 25k \le 40 \implies k \le 1 \). (1 triplet)

For \( r = \frac{5}{3} \): Triplet \( (9k, 15k, 25k) \). \( 25k \le 40 \implies k \le 1 \). (1 triplet)

For \( r = \frac{5}{4} \): Triplet \( (16k, 20k, 25k) \). \( 25k \le 40 \implies k \le 1 \). (1 triplet)

For \( r = \frac{6}{5} \): Triplet \( (25k, 30k, 36k) \). \( 36k \le 40 \implies k \le 1 \). (1 triplet)

Total for non-integer ratios: \( 4 + 2 + 1 + 1 + 1 + 1 = 10 \).

Step 5: Compute the total number of favorable outcomes and the probability.

The total count of favorable outcomes is the sum from both cases:

\[ N_{\text{favorable}} = 18 + 10 = 28 \]

The probability is calculated as:

\[ P = \frac{N_{\text{favorable}}}{N_{\text{total}}} = \frac{28}{9880} \]

Simplify the fraction:

\[ \frac{28}{9880} = \frac{7}{2470} \]

Since 7 is prime and 2470 is not divisible by 7 (\( 2470 = 7 \times 352 + 6 \)), the fraction \( \frac{7}{2470} \) is in its simplest form. Thus, \( m = 7 \) and \( n = 2470 \).

Final Calculation:

The required value is \( m + n \).

\[ m + n = 7 + 2470 = 2477 \]

The value of \( m + n \) is 2477.