Competition · AMC preparation · step 4 of 4
AMC 8 · 2001 · #25
Grade 5 number-theoryPick an answer.
AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Bound the multiplier
Every permutation lies in [2457, 7542], so k = N/d ＜ 4; since k ≠ 1, only k = 2 or k = 3 can work.
Because every permutation lies in the same narrow band [2457, 7542], one such number cannot be more than three times another.
If one arrangement of the digits 2, 4, 5, 7 is a whole-number multiple of another such arrangement, the multiplier can only be 2 or 3.
▸ Why?
The multiplier is a whole number that is at least 2, because one arrangement is a multiple of a different arrangement.
▸ Why?
Calling one number a multiple of another means it equals that other number taken some whole number of times.
▸ Why?
That whole number of times cannot be 1, since taking a number one time leaves it unchanged, yet here the two arrangements are different.
▸ Why?
The multiplier is at most 3, because four copies of even the smallest arrangement already exceed the largest arrangement.
▸ Why?
The smallest arrangement is 2457 and the largest is 7542, because a digit counts for more when it sits in a higher place.
▸ Why?
Four times 2457 is 9828, which is bigger than 7542; comparing the two totals place by place from the thousands down shows 9828 is larger, so no arrangement can be four or more times another.
Test choice A
Test (A) 5724: 5724 ÷ 2 = 2862 and 5724 ÷ 3 = 1908 both include forbidden digits, so drop it.
Neither quotient is a rearrangement of {2, 4, 5, 7}, so 5724 has no qualifying divisor.
5.NBT.B.6Eliminate PossibilitiesTest choice B
Test (B) 7245: it is odd so ÷ 2 fails, and 7245 ÷ 3 = 2415 sneaks in a 1 — eliminate.
An odd number can't be twice an integer, and the cube-friendly quotient 2415 uses the wrong digits.
5.NBT.B.6Eliminate PossibilitiesTest choice C
Test (C) 7254: 7254 ÷ 2 = 3627 and 7254 ÷ 3 = 2418 both bring in stray digits — eliminate.
Both quotients introduce digits outside {2, 4, 5, 7}, so 7254 also has no qualifying divisor.
5.NBT.B.6Eliminate PossibilitiesTest choice D
Test (D) 7425: it is odd, and 7425 ÷ 3 = 2475 is again a permutation of 2, 4, 5, 7 — keep it.
2475 is one of the 24 permutations, so 7425 = 3 × 2475 is exactly the "multiple of another one" the problem describes.
4.OA.B.4Eliminate PossibilitiesTest choice E
Test (E) 7542: 7542 ÷ 2 = 3771 and 7542 ÷ 3 = 2514 each carry a forbidden digit — eliminate.
Neither quotient lands on a permutation, leaving (D) as the unique survivor.
5.NBT.B.6Eliminate PossibilitiesRead off the answer
Only (D) survives: 7425 = 3 × 2475, the one permutation that is a multiple of another.
Listing all candidate-by-divisor checks leaves exactly one valid pair, which is the answer.
4.OA.B.4Make A Systematic ListEvery 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).
- Bound the multiplier
- Test choice A
- Test choice B
- Test choice C
- Test choice D
- Test choice E
- Read off the answer
A parent dashboard for the family lives at sensimlab.com.