Question

Three distinct natural numbers are selected from 1 to 10. The probability that they are in geometric progression is $\frac{p}{q}$, where $p$ and $q$ are relatively prime. Then, the number of prime divisors of $pq$ is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (B)

3

Answer 2

The question asks for the number of prime divisors of $pq$, where $\frac{p}{q}$ is the probability of selecting three distinct natural numbers from 1 to 10 that form a geometric progression.

1. Total possible combinations: The total number of ways to choose 3 distinct numbers from 1 to 10 is given by the combination formula:

   $\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

2. Favorable outcomes (geometric progressions): We need to find triples $(a, b, c)$ such that $b^2 = ac$ and $1 \le a, b, c \le 10$. Listing the possible geometric progressions:

   • $b = 2$: $(1, 2, 4)$
   • $b = 3$: $(1, 3, 9)$
   • $b = 4$: $(2, 4, 8)$
   • $b = 6$: $(4, 6, 9)$

   There are 4 such triples.

3. Probability calculation: The probability of selecting a geometric progression is:

   $\frac{4}{120} = \frac{1}{30}$

   Thus, $p = 1$ and $q = 30$.

4. Finding the prime divisors of $pq$: The product $pq = 1 \times 30 = 30$. The prime factorization of 30 is $2 \times 3 \times 5$. Therefore, the prime divisors of 30 are 2, 3, and 5.

5. Final answer: The number of prime divisors of $pq$ is 3.