Problem 3 · 2022 AMC 8
Medium
Number Theory
factorizationfactor-triplescasework
When three positive integers a, b, and c are multiplied together, their product is 100. Suppose a < b < c. In how many ways can the numbers be chosen?
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Answer: E — 4 ways.
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Hint 1 of 2
Because a < b < c, the smallest factor can't be big — if all three were that size the product would blow past 100. So fix a first; it has very few options.
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Hint 2 of 2
a must be a small factor of 100 (it's the least of three, so a³ < 100 ⇒ a ≤ 4). For each such a, split the leftover product into b < c.
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Approach: fix the smallest factor and list outward
- Insight: since a < b < c, the smallest factor a is the natural thing to pin down — it's tiny (if all three were as big as a, the product a³ would already exceed 100). So a can only be a small factor of 100.
- a = 1: then bc = 100 with b < c — (1,2,50), (1,4,25), (1,5,20).
- a = 2: then bc = 50 with 2 < b < c — only (2,5,10).
- Bigger a leaves no room (e.g. a = 4 would need bc = 25 with 4 < b < c, impossible). That gives 4 valid triples.
- Why fix the smallest: in any “a < b < c, fixed product” problem, the smallest is squeezed into a short list, so looping over it is the fast, miss-nothing path.
Another way — bound the smallest factor first (MAA):
- Since a < b < c and abc = 100, we get a3 < 100, so a ≤ 4. And a must be a factor of 100, so a ∈ {1, 2, 4}.
- For each: a = 1 gives (1,2,50), (1,4,25), (1,5,20); a = 2 gives (2,5,10); a = 4 gives nothing (would need bc = 25 with 4 < b < c).
- Total: 4 ways.
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