Problem 16 · 2009 AMC 8
Medium
Counting & Probability
factor-triples
How many 3-digit positive integers have digits whose product equals 24?
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Answer: D — 21.
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Hint 1 of 2
Split the job in two: FIRST find which sets of three digits multiply to 24 (order ignored), THEN count how many numbers each set makes by reordering. Don't try to do both at once.
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Hint 2 of 2
Each digit must be 1–9 (no 0, or the product is 0). To list the sets cleanly, factor 24 = 2³ × 3 and hand the factors out three ways. Watch for repeats — a set with a doubled digit has fewer arrangements.
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Approach: list digit-sets, then count orderings (correct for repeats)
- Digit-sets (each 1–9) with product 24: {1,3,8}, {1,4,6}, {2,3,4}, {2,2,6}. That's it — any other split needs a digit above 9.
- Count orderings of each. Three different digits give 3! = 6 numbers each: that's 6 + 6 + 6 = 18 from the first three sets.
- {2,2,6} has a repeated 2, so swapping the two 2's gives nothing new: only 3!/2! = 3 numbers.
- Total: 18 + 3 = 21.
- Why this transfers: "count the numbers with property X" usually splits into "find the unordered ingredient sets, then count arrangements." And always divide out duplicate items — forgetting the repeat is the #1 overcount error.
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