AMC 8 · 2001 · #25

Grade 5 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

divisibility-rulesmulti-digit-arithmeticdigit-constraintssystematic-enumeration systematic-enumerationdigit-constraintsguess-and-check ↑ Prerequisites: divisibility-rulesmulti-digit-arithmetic

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Problem

There are 24 four-digit numbers that use each of the digits 2, 4, 5, 7 exactly once. Exactly one of these 24 numbers is a multiple of another number on that same list. Among the five answer choices, find that special number.

Pick an answer.

(A)

5724

(B)

7245

(C)

7254

(D)

7425

(E)

7542

AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

Only five candidates are offered, so Tool #3 (Eliminate Possibilities) is the natural lead: divide each candidate by a small integer and check whether the quotient is also a permutation of {2, 4, 5, 7}. Tool #5 (Look for a Pattern) narrows what divisors we even need to try: every permutation has digit sum 2+4+5+7 = 18, so every permutation is divisible by 9 — and because the smallest permutation is 2457 and the largest is 7542, the multiplier k = N/d can only be 2 or 3. Tool #2 (Make a Systematic List) keeps the at most ten quotient checks tidy. We avoid Tool #13 (Algebra) because there is nothing to solve symbolically — divisibility filtering does it all.

1STEP 1

Every permutation lies in [2457, 7542], so k = N/d ＜ 4; since k ≠ 1, only k = 2 or k = 3 can work.

k = N/d ≤

\frac{7542}{2457}

＜ 4 → k ∈ {2, 3}

2STEP 2

Test (A) 5724: 5724 ÷ 2 = 2862 and 5724 ÷ 3 = 1908 both include forbidden digits, so drop it.

5724 ÷ 2 = 2862 (has 8, 6, not allowed); 5724 ÷ 3 = 1908 (has 1, 9, 0, 8, not allowed). Eliminate (A).

3STEP 3

Test (B) 7245: it is odd so ÷ 2 fails, and 7245 ÷ 3 = 2415 sneaks in a 1 — eliminate.

7245 ÷ 3 = 2415 (digits 2, 4, 1, 5 include a 1, not allowed). Eliminate (B).

4STEP 4

Test (C) 7254: 7254 ÷ 2 = 3627 and 7254 ÷ 3 = 2418 both bring in stray digits — eliminate.

7254 ÷ 2 = 3627 (has 3, 6, not allowed); 7254 ÷ 3 = 2418 (has 1, 8, not allowed). Eliminate (C).

5STEP 5

Test (D) 7425: it is odd, and 7425 ÷ 3 = 2475 is again a permutation of 2, 4, 5, 7 — keep it.

7425 ÷ 3 = 2475. The digits of 2475 are {2, 4, 7, 5} — a permutation of the original set. Keep (D).

6STEP 6

Test (E) 7542: 7542 ÷ 2 = 3771 and 7542 ÷ 3 = 2514 each carry a forbidden digit — eliminate.

7542 ÷ 2 = 3771 (has 3, 1, not allowed); 7542 ÷ 3 = 2514 (has 1, not allowed). Eliminate (E).

7STEP 7

Only (D) survives: 7425 = 3 × 2475, the one permutation that is a multiple of another.

7425 = 3 × 2475 → (D)

Answer

7425

Double-check the keeper. 3 × 2475 = 3 × 2000 + 3 × 475 = 6000 + 1425 = 7425. The digits of 2475 are 2, 4, 7, 5 — exactly the required set, used once each — and 2475 is indeed one of the 24 permutations. Also, both numbers have digit sum 18, so each is divisible by 9, which is consistent with one being three times the other. None of the other four choices produced a valid permutation quotient, so the uniqueness claim in the problem is matched: exactly one of the 24 permutations is a multiple of another, and it is 7425.

💡Key takeaway

Every rearrangement of {2, 4, 5, 7} sits between 2457 and 7542, so one can only be at most 3 times another. Testing ÷ 2 and ÷ 3 on each of the five choices, the only quotient that comes back as a permutation is 7425 ÷ 3 = 2475 — answer (D).

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