Question: Three distinct prime numbers less than 100 are selected. What is the probability that their product is divisible by 15? - RoadRUNNER Motorcycle Touring & Travel Magazine

Understanding the Context

Public interest in math has surged in recent years, fueled by viral content on platforms emphasizing logic, puzzles, and statistical reasoning. The exploration of primes, modular arithmetic, and divisibility tests stands out in digital conversations—especially among users seeking intellectual challenge without emotional or explicit content. The question stems from a natural curiosity: how do basic number properties play out probabilistically? This blend of simplicity and depth makes it ideal for automated content aiming to capture attention deeply. On mobile-driven Discover feeds, clarity and precision boost discoverability, turning abstract logic into shareable insight.

Why This Question Is Gaining Attention in the US

In an era where financial literacy, data understanding, and pattern recognition are prioritized, the question taps into a growing culture of self-directed learning. From STEM engagement to algorithmic thinking, Americans are increasingly comfortable questioning underlying mechanics behind numbers and systems. While the topic avoids controversial territory, its structure invites thoughtful exploration—an appeal that aligns with natural curiosity about cause, effect, and chance within bounded parameters like “three distinct primes under 100.” Mobile users benefit from concise, scannable explanations that reward careful reading, increasing dwell time and engagement.

Key Insights

Exactly What This Question Explains—Building from the Basics

At its core, determining whether the product of three distinct primes below 100 is divisible by 15 requires understanding divisibility rules. A number is divisible by 15 if and only if it is divisible by both 3 and 5—prime factors that demand specific inclusion in the multiplicands. Among all primes under 100 (there are 25 total), only two qualify: 3 and 5. Any valid triplet must include both to yield a product divisible by 15. So the task reduces to computing the probability that both 3 and 5 are selected when picking three distinct primes.

To answer this, we begin by calculating:

• Total number of ways to choose 3 distinct primes from 25:
  [ \binom{25}{3} = \frac{25 \cdot 24 \cdot 23}{6} = 2300 ]

Final Thoughts

• Favorable selections: triplets that include both 3 and 5, plus one additional prime from the remaining