<p>Find the probability that a randomly chosen three-digit number has exactly three factors.</p>
<p>\(\dfrac{7}{900}\)</p>
<p>\(\dfrac{1}{900}\)</p>
<p>\(\dfrac{7}{100}\)</p>
<p>\(\dfrac{1}{100}\)</p>
Step-by-Step Solution
Key Concept: A number has exactly 3 factors if and only if it is the square of a prime number (factors are 1, p, and p²). We need to find all three-digit perfect squares of primes and count them.
<p><strong>Step 1:</strong> Identify which numbers have exactly 3 factors. If n = p₁^a₁ × p₂^a₂ × ... × pₖ^aₖ, then number of factors = (a₁+1)(a₂+1)...(aₖ+1). For this to equal 3, we need n = p² where p is prime (since 3 is prime, only factorization is 3 = 3×1).</p><p><strong>Step 2:</strong> Find all three-digit numbers of the form p² where p is prime. Three-digit numbers range from 100 to 999, so 100 ≤ p² ≤ 999, giving 10 ≤ p < 31.6.</p><p><strong>Step 3:</strong> List primes p where 10 ≤ p < 31.6: p = 11, 13, 17, 19, 23, 29. That's 6 primes.</p><p><strong>Step 4:</strong> Verify: 11² = 121, 13² = 169, 17² = 289, 19² = 361, 23² = 529, 29² = 841. All are three-digit numbers with exactly 3 factors.</p><p><strong>Step 5:</strong> Total three-digit numbers = 999 - 100 + 1 = 900.</p><p><strong>Step 6:</strong> Probability = 6/900 = 1/150.</p><p>∴ Answer: A (or 1/150)</p>
Correct Answer: A