AMC 8 · 2001 · #25
Easy mode Grade 5Problem
Take the four digits 2, 4, 5, and 7. Using each digit exactly once, you can make 24 different four-digit numbers (like 2457, 2475, 2547, and so on).
Out of those 24 numbers, exactly one of them is a multiple of another one in the same list. In other words, one number on the list divides evenly into another number on the list.
Which of the following is that special number?
Pick an answer.
AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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.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 $k = 3$.
💡 Because every permutation lies in the same narrow band $[2457, 7542]$, one such number cannot be more than three times another.
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 $\{2, 4, 5, 7\}$.
💡 Neither quotient is a rearrangement of $\{2, 4, 5, 7\}$, so $5724$ has no qualifying divisor.
5.NBT.B.6 Step 3 - Test choice (B) $7245$.
- It is odd, so $k = 2$ fails automatically.
- Try $k = 3$.
💡 An odd number can't be twice an integer, and the cube-friendly quotient $2415$ uses the wrong digits.
5.NBT.B.6 Step 4 - Test choice (C) $7254$.
- Try both $k = 2$ and $k = 3$.
💡 Both quotients introduce digits outside $\{2, 4, 5, 7\}$, so $7254$ also has no qualifying divisor.
4.OA.B.4 Step 5 - Test choice (D) $7425$.
- It is odd, so try $k = 3$.
💡 $2475$ is one of the $24$ permutations, so $7425 = 3 \times 2475$ is exactly the "multiple of another one" the problem describes.
5.NBT.B.6 Step 6 Test choice (E) $7542$ to confirm only (D) works.
💡 Neither quotient lands on a permutation, leaving (D) as the unique survivor.
4.OA.B.4 Step 7 - Read off the answer.
- Only choice (D) passes the divisibility test, and the matching pair is $7425 = 3 \times 2475$.
💡 Listing all candidate-by-divisor checks leaves exactly one valid pair, which is the answer.
5.NBT.A.3 Bound the multiplier $k$. If $N = k \cdot d$ with both $N$ and $d$ on the list, 5.NBT.B.6 Test choice (A) $5724$. Divide by $2$ and by $3$ and check whether the quotient 5.NBT.B.6 Test choice (B) $7245$. It is odd, so $k = 2$ fails automatically. Try $k = 3$. 5.NBT.B.6 Test choice (C) $7254$. Try both $k = 2$ and $k = 3$. 4.OA.B.4 Test choice (D) $7425$. It is odd, so try $k = 3$. 5.NBT.B.6 Test choice (E) $7542$ to confirm only (D) works. 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.4Find 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.6Find 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.3Read, 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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