Question: Suppose that $ v $ is a positive multiple of 5. If $ v $ cubed is less than 2000, what is the greatest possible value of $ v $?

Sat, 02 May 2026 14:59:31 +0000

**What’s the Greatest Multiple of 5 Whose Cube Stays Under 2000? A Clear, Practical Guide** In a world increasingly shaped by numerical puzzles and key thresholds, a simple math question gains quiet relevance: Suppose that $ v $ is a positive multiple of 5. If $ v^3 $ is less than 2000, what’s the highest value $ v $ can reach? This query reflects a common curiosity—how small, whole-number multiples stack up under real-world limits. Understanding the answer supports smarter decision-making in education, project planning, and digital goal-setting across the U.S. The math behind this question reveals clear boundaries. Compounding the cumulating impact of triple-digit cubes, numbers like 5, 10, and 15 scale rapidly. $ 5^3 = 125 $, $ 10^3 = 1,000 $, but $ 15^3 = 3,375 $, already overtopping 2,000. So the cube of the next multiple, 15, exceeds the threshold—but 10 stays safely below. This pattern holds: $ 10^3 = 1,000 $ is under 2,000, while $ 15^3 = 3,375 $ crosses it. **Answering the Core Question: The Strong Evidence Points to $ v = 10 $**
The largest multiple of 5 whose cube remains under 2,000 is $ v = 10 $. At this level, $ v^3 = 1,000 $—well below the 2,000 limit. Moving to $ v = 15$, the cube jumps to 3,375, too high. Thus, 10 stands as the maximum, confirmed by structured testing and clear numerical boundaries.