Problem 25 · 2001 AMC 8
Stretch
Number Theory
divisibilitycareful-counting
There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5, and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?
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Answer: D — 7425.
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Hint 1 of 2
Both the number and its factor-partner are built from the same four digits, so both live in the narrow window 2457 to 7542 — that squeezes the possible multiplier down to almost nothing.
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Hint 2 of 2
Since even 2457 × 4 ≈ 9828 already escapes the window, the factor can only be 2 or 3. Test small.
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Approach: the only feasible factor is 3
- Both numbers use exactly {2, 4, 5, 7}, so each lies between 2457 and 7542. For one to be a multiple of the other, the ratio must keep the product inside that window — and 2457 × 4 ≈ 9828 already overshoots, so the multiplier is just 2 or 3.
- Doubling never lands back in the set, so try ×3: 2475 × 3 = 7425, which uses exactly 2, 4, 5, 7. ✓
- So the answer is 7425 (= 3 × 2475). Neat confirmation: every arrangement of 2,4,5,7 has digit sum 2+4+5+7 = 18, a multiple of 9 — so all 24 numbers are multiples of 9, and 7425 = 9 × 825 fits right in.
- The reusable idea: when a multiple must reuse a fixed digit set, the size window pins the factor to a tiny range — bounding before searching turns 24 candidates into one quick test.
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