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AMC 8 · 2001 · #25

Grade 5 number-theory
Archetype Small-Domain Constrained Search
divisibility-rulesmulti-digit-arithmeticdigit-constraintssystematic-enumeration systematic-enumerationdigit-constraintsguess-and-check ↑ Prerequisites: divisibility-rulesmulti-digit-arithmetic
📏 Short solution 💡 3 insights
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Problem

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?

Pick an answer.

(A)
5724
(B)
7245
(C)
7254
(D)
7425
(E)
7542
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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Each four-digit number uses the digits $2, 4, 5, 7$ exactly once; There are $24 = 4!$ such numbers in total; Exactly one of them is a multiple of another one (also from the $24$); Answer choices: (A) $5724$, (B) $7245$, (C) $7254$, (D) $7425$, (E) $7542$

Unknowns: Which choice is a multiple of some other permutation of $\{2, 4, 5, 7\}$

Understand

Restated: 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.

Givens: Each four-digit number uses the digits $2, 4, 5, 7$ exactly once; There are $24 = 4!$ such numbers in total; Exactly one of them is a multiple of another one (also from the $24$); Answer choices: (A) $5724$, (B) $7245$, (C) $7254$, (D) $7425$, (E) $7542$

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #5 Look for a Pattern, #2 Make a Systematic List

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.

Execute — Answer: D

#5 Look for a Pattern 5.NBT.A.3 Step 1
  • Bound the multiplier $k$.
  • If $N = k \cdot d$ with both $N$ and $d$ on the list, then $d \geq 2457$ (smallest permutation) and $N \leq 7542$ (largest).
  • So $k = N/d \leq 7542/2457 < 4$.
  • Since $k \neq 1$, the only possibilities are $k = 2$ or $k = 3$.
$$k = \dfrac{N}{d} \leq \dfrac{7542}{2457} < 4 \;\Rightarrow\; k \in \{2, 3\}$$

💡 Because every permutation lies in the same narrow band $[2457, 7542]$, one such number cannot be more than three times another.

#3 Eliminate Possibilities 5.NBT.B.6 Step 2
  • Test choice (A) $5724$.
  • Divide by $2$ and by $3$ and check whether the quotient uses exactly the digits $\{2, 4, 5, 7\}$.
$5724 \div 2 = 2862$ (has $8, 6$, not allowed); $\;5724 \div 3 = 1908$ (has $1, 9, 0, 8$, not allowed). Eliminate (A).

💡 Neither quotient is a rearrangement of $\{2, 4, 5, 7\}$, so $5724$ has no qualifying divisor.

#3 Eliminate Possibilities 5.NBT.B.6 Step 3
  • Test choice (B) $7245$.
  • It is odd, so $k = 2$ fails automatically.
  • Try $k = 3$.
$7245 \div 3 = 2415$ (digits $2, 4, 1, 5$ include a $1$, not allowed). Eliminate (B).

💡 An odd number can't be twice an integer, and the cube-friendly quotient $2415$ uses the wrong digits.

#3 Eliminate Possibilities 5.NBT.B.6 Step 4
  • Test choice (C) $7254$.
  • Try both $k = 2$ and $k = 3$.
$7254 \div 2 = 3627$ (has $3, 6$, not allowed); $\;7254 \div 3 = 2418$ (has $1, 8$, not allowed). Eliminate (C).

💡 Both quotients introduce digits outside $\{2, 4, 5, 7\}$, so $7254$ also has no qualifying divisor.

#3 Eliminate Possibilities 4.OA.B.4 Step 5
  • Test choice (D) $7425$.
  • It is odd, so try $k = 3$.
$7425 \div 3 = 2475$. The digits of $2475$ are $\{2, 4, 7, 5\}$ — a permutation of the original set. Keep (D).

💡 $2475$ is one of the $24$ permutations, so $7425 = 3 \times 2475$ is exactly the "multiple of another one" the problem describes.

#3 Eliminate Possibilities 5.NBT.B.6 Step 6

Test choice (E) $7542$ to confirm only (D) works.

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

💡 Neither quotient lands on a permutation, leaving (D) as the unique survivor.

#2 Make a Systematic List 4.OA.B.4 Step 7
  • Read off the answer.
  • Only choice (D) passes the divisibility test, and the matching pair is $7425 = 3 \times 2475$.
$$7425 = 3 \times 2475 \;\Rightarrow\; \textbf{(D)}$$

💡 Listing all candidate-by-divisor checks leaves exactly one valid pair, which is the answer.

[1] #5 5.NBT.A.3 Bound the multiplier $k$. If $N = k \cdot d$ with both $N$ and $d$ on the list,
[2] #3 5.NBT.B.6 Test choice (A) $5724$. Divide by $2$ and by $3$ and check whether the quotient
[3] #3 5.NBT.B.6 Test choice (B) $7245$. It is odd, so $k = 2$ fails automatically. Try $k = 3$.
[4] #3 5.NBT.B.6 Test choice (C) $7254$. Try both $k = 2$ and $k = 3$.
[5] #3 4.OA.B.4 Test choice (D) $7425$. It is odd, so try $k = 3$.
[6] #3 5.NBT.B.6 Test choice (E) $7542$ to confirm only (D) works.
[7] #2 4.OA.B.4 Read off the answer. Only choice (D) passes the divisibility test, and the match

Review

Reasonableness: Double-check the keeper. $3 \times 2475 = 3 \times 2000 + 3 \times 475 = 6000 + 1425 = 7425$. 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 $7425$.

Alternative: Tool #5 (Look for a Pattern) shortcut: every permutation has digit sum $18$, so every permutation is a multiple of $9$. That alone limits the multiplier to $k \in \{2, 3\}$ (as bounded above). For $k = 3$, the divisor $d$ must satisfy $3d \leq 7542$, so $d \leq 2514$ — only permutations starting with $24$ or $25$ qualify, namely $2457, 2475, 2547, 2574$. Compute $3d$ for each: $3 \cdot 2457 = 7371$, $3 \cdot 2475 = 7425$, $3 \cdot 2547 = 7641$, $3 \cdot 2574 = 7722$. Only $7425$ is itself a permutation, confirming (D) without checking $k = 2$ at all.

CCSS standards used (min grade 5)

  • 4.OA.B.4 Find factor pairs and recognize multiples of a whole number (Reading $7425 = 3 \times 2475$ as the multiple-divisor pair the problem asks for, and recognizing the other four choices have no valid factor of this kind on the permutation list.)
  • 5.NBT.B.6 Find whole-number quotients with up to four-digit dividends (Computing each quotient $N \div 2$ and $N \div 3$ for the five answer choices and checking which digits appear in the result.)
  • 5.NBT.A.3 Read, write, and compare decimals using place-value understanding (Comparing magnitudes (smallest permutation $2457$, largest $7542$) to bound the possible multiplier $k$ at $k \leq 3$.)

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

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

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