1. Key Concepts and Formulas

• Combinations: The number of ways to choose $k$ distinct items from a set of $n$ distinct items, where the order of selection does not matter, is given by the combination formula: $^nC_k = \binom{n}{k} = \frac{n!}{k!(n-k)!}$
• Geometric Progression (G.P.): A sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For three numbers $a, b, c$ to be in G.P., the condition $b^2 = ac$ must hold. For an increasing G.P., the terms must satisfy $a < b < c$, which implies the common ratio $r$ must be greater than 1 ($r > 1$).
• Probability: The probability of an event $E$ is the ratio of the number of favorable outcomes to the total number of possible outcomes: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

2. Step-by-Step Solution

Step 1: Calculate the total number of possible outcomes. We need to select three distinct numbers randomly from the set $S = \{1, 2, 3, \ldots, 40\}$. Since the order of selection does not matter for forming a set of three numbers, we use combinations. The total number of ways to choose 3 distinct numbers from 40 is: $\text{Total Outcomes} = ^{40}C_3 = \frac{40!}{3!(40-3)!} = \frac{40 \times 39 \times 38}{3 \times 2 \times 1}$ $\text{Total Outcomes} = 40 \times 13 \times 19 = 520 \times 19 = 9880$

Step 2: Determine the structure of an increasing G.P. with integer terms. Let the three distinct numbers in an increasing G.P. be $a, b, c$. Since they are in G.P., $b = ar$ and $c = br = ar^2$ for some common ratio $r$. Since $a, b, c$ are distinct and increasing, we must have $r > 1$. Also, $a, b, c$ must be integers from the given set. This implies $r$ must be a rational number. Let $r = \frac{p}{q}$, where $p, q$ are coprime positive integers with $p > q \ge 1$. The terms of the G.P. are $a, a\left(\frac{p}{q}\right), a\left(\frac{p}{q}\right)^2$. For $a\left(\frac{p}{q}\right)$ to be an integer, $q$ must divide $a$. For $a\left(\frac{p}{q}\right)^2 = a\left(\frac{p^2}{q^2}\right)$ to be an integer, $q^2$ must divide $a$ (since $\operatorname{gcd}(p, q) = 1$, it follows that $\operatorname{gcd}(p^2, q^2) = 1$). So, $a$ must be a multiple of $q^2$. Let $a = k q^2$ for some positive integer $k$. Then the three numbers are: $k q^2, \quad k q^2 \left(\frac{p}{q}\right), \quad k q^2 \left(\frac{p}{q}\right)^2$ $k q^2, \quad k p q, \quad k p^2$ These three numbers must be distinct, positive, and less than or equal to 40. So, we need to find integer values for $k, p, q$ such that $1 \le k q^2 < k p q < k p^2 \le 40$, with $p > q \ge 1$ and $\operatorname{gcd}(p, q) = 1$.

Step 3: List all possible increasing G.P.'s (favorable outcomes). We systematically iterate through possible values for $q$, $p$, and $k$. The condition $k p^2 \le 40$ will limit the values.

• Case 1: $q=1$ (integer common ratio $r=p$) The G.P. terms are $k, kp, kp^2$. Condition: $kp^2 \le 40$.

  • If $p=2$: $k, 2k, 4k$. $4k \le 40 \implies k \le 10$. (10 G.P.'s) Examples: $(1,2,4), (2,4,8), \ldots, (10,20,40)$.
  • If $p=3$: $k, 3k, 9k$. $9k \le 40 \implies k \le 4$. (4 G.P.'s) Examples: $(1,3,9), (2,6,18), (3,9,27), (4,12,36)$.
  • If $p=4$: $k, 4k, 16k$. $16k \le 40 \implies k \le 2$. (2 G.P.'s) Examples: $(1,4,16), (2,8,32)$.
  • If $p=5$: $k, 5k, 25k$. $25k \le 40 \implies k \le 1$. (1 G.P.) Example: $(1,5,25)$.
  • If $p=6$: $k, 6k, 36k$. $36k \le 40 \implies k \le 1$. (1 G.P.) Example: $(1,6,36)$.
  • If $p \ge 7$: $kp^2 \ge 1 \cdot 7^2 = 49 > 40$. No more G.P.'s. Total for $q=1$: $10 + 4 + 2 + 1 + 1 = 18$ G.P.'s.

• Case 2: $q=2$ (common ratio $r=p/2$, $p$ must be odd and $p>2$) The G.P. terms are $4k, 2kp, kp^2$. Condition: $kp^2 \le 40$.

  • If $p=3$: $4k, 6k, 9k$. $9k \le 40 \implies k \le 4$. (4 G.P.'s) Examples: $(4,6,9), (8,12,18), (12,18,27), (16,24,36)$.
  • If $p=5$: $4k, 10k, 25k$. $25k \le 40 \implies k \le 1$. (1 G.P.) Example: $(4,10,25)$.
  • If $p \ge 7$: $kp^2 \ge 1 \cdot 7^2 = 49 > 40$. No more G.P.'s. Total for $q=2$: $4 + 1 = 5$ G.P.'s.

• Case 3: $q=3$ (common ratio $r=p/3$, $p$ not a multiple of 3 and $p>3$) The G.P. terms are $9k, 3kp, kp^2$. Condition: $kp^2 \le 40$.

  • If $p=4$: $9k, 12k, 16k$. $16k \le 40 \implies k \le 2$. (2 G.P.'s) Examples: $(9,12,16), (18,24,32)$.
  • If $p=5$: $9k, 15k, 25k$. $25k \le 40 \implies k \le 1$. (1 G.P.) Example: $(9,15,25)$.
  • If $p \ge 7$: $kp^2 \ge 1 \cdot 7^2 = 49 > 40$. No more G.P.'s. Total for $q=3$: $2 + 1 = 3$ G.P.'s.

• Case 4: $q=4$ (common ratio $r=p/4$, $p$ must be odd and $p>4$) The G.P. terms are $16k, 4kp, kp^2$. Condition: $kp^2 \le 40$.

  • If $p=5$: $16k, 20k, 25k$. $25k \le 40 \implies k \le 1$. (1 G.P.) Example: $(16,20,25)$.
  • If $p \ge 7$: $kp^2 \ge 1 \cdot 7^2 = 49 > 40$. No more G.P.'s. Total for $q=4$: $1$ G.P.

• Case 5: $q=5$ (common ratio $r=p/5$, $p$ not a multiple of 5 and $p>5$) The G.P. terms are $25k, 5kp, kp^2$. Condition: $kp^2 \le 40$.

  • If $p=6$: $25k, 30k, 36k$. $36k \le 40 \implies k \le 1$. (1 G.P.) Example: $(25,30,36)$. (Note: $\operatorname{gcd}(6,5)=1$ is true).
  • If $p \ge 7$: $kp^2 \ge 1 \cdot 7^2 = 49 > 40$. No more G.P.'s. Total for $q=5$: $1$ G.P.

• Case 6: $q \ge 6$ If $q=6$, the smallest $p$ such that $p>6$ and $\operatorname{gcd}(p,6)=1$ is $p=7$. The G.P. terms would be $36k, 42k, 49k$. $49k \le 40 \implies$ no possible integer $k$. Thus, no G.P.'s for $q \ge 6$.

Total number of favorable outcomes (increasing G.P.'s) = $18 + 5 + 3 + 1 + 1 = 28$.

Step 4: Calculate the probability. $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{28}{9880}$

Step 5: Simplify the probability and find $m+n$. We need to simplify the fraction $\frac{28}{9880}$ to its simplest form $\frac{m}{n}$ where $\operatorname{gcd}(m, n)=1$. Both 28 and 9880 are divisible by 4: $\frac{28}{9880} = \frac{28 \div 4}{9880 \div 4} = \frac{7}{2470}$ Now we check if $\operatorname{gcd}(7, 2470)=1$. The prime factorization of 7 is just 7. To check if 2470 is divisible by 7: $2470 \div 7 \approx 352.85$, so 2470 is not divisible by 7. Therefore, $\operatorname{gcd}(7, 2470)=1$. So, $m=7$ and $n=2470$. Finally, we need to find $m+n$: $m+n = 7 + 2470 = 2477$

3. Common Mistakes & Tips

• Missing Rational Ratios: A common mistake is to only consider integer common ratios ($q=1$). Remember that G.P. terms can be integers even if the common ratio is a fraction (e.g., $4, 6, 9$ has $r=3/2$).
• Incorrectly Calculating Total Outcomes: Ensure you use combinations ($^nC_k$) for selecting distinct items when order doesn't matter, as is the case here for forming a set of three numbers.
• Forgetting Coprime Condition: When representing the common ratio as $p/q$, always ensure $p$ and $q$ are coprime. This prevents double-counting (e.g., $r=2/1$ and $r=4/2$ would generate the same sequences if $q$ wasn't reduced to its simplest form).

4. Summary

The problem required us to calculate the probability that three distinct numbers chosen randomly from $\{1, 2, \ldots, 40\}$ form an increasing Geometric Progression. We first determined the total number of ways to select three distinct numbers using combinations, which was 9880. Next, we systematically listed all possible integer G.P.'s within the given range by representing the terms as $kq^2, kpq, kp^2$, where $p/q$ is the common ratio in simplest form. We found 28 such G.P.'s. The probability was then calculated as the ratio of favorable outcomes to total outcomes, $\frac{28}{9880}$, which simplified to $\frac{7}{2470}$. Thus, $m=7$ and $n=2470$, and their sum $m+n$ is 2477.

5. Final Answer

The final answer is $\boxed{2477}$.