AMC 8 · 2011 · #23
Grade 5 countingProblem
How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?
Pick an answer.
AMC 8 2011 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: Count the 4-digit positive integers $\overline{ABCD}$ whose four digits are all different, with leading digit $A \neq 0$, the whole number divisible by $5$, and the digit $5$ being the largest digit used.
Givens: The integer has exactly $4$ digits $A, B, C, D$ and they must all be distinct; $A \neq 0$ (no leading zero); The integer is a multiple of $5$, so $D \in \{0, 5\}$; $5$ is the largest digit used, so every digit lies in $\{0, 1, 2, 3, 4, 5\}$ and the digit $5$ appears; Answer choices: (A) $24$, (B) $48$, (C) $60$, (D) $84$, (E) $108$
Unknowns: The total count of 4-digit integers $\overline{ABCD}$ that satisfy all four conditions
Understand
Restated: Count the 4-digit positive integers $\overline{ABCD}$ whose four digits are all different, with leading digit $A \neq 0$, the whole number divisible by $5$, and the digit $5$ being the largest digit used.
Givens: The integer has exactly $4$ digits $A, B, C, D$ and they must all be distinct; $A \neq 0$ (no leading zero); The integer is a multiple of $5$, so $D \in \{0, 5\}$; $5$ is the largest digit used, so every digit lies in $\{0, 1, 2, 3, 4, 5\}$ and the digit $5$ appears; Answer choices: (A) $24$, (B) $48$, (C) $60$, (D) $84$, (E) $108$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
The divisibility-by-$5$ rule forces $D = 0$ or $D = 5$, so Tool #7 (Identify Subproblems) splits the count into two clean cases on the last digit. Inside each case, Tool #2 (Make a Systematic List) counts the choices position by position using the multiplication principle: pick $A$, then $B$, then $C$ from the digits that remain in $\{0,1,2,3,4,5\}$. Tool #3 (Eliminate Possibilities) handles the two side-conditions — $A \neq 0$ removes one option from the thousands slot, and "$5$ must appear somewhere" forces a placement choice in the $D = 0$ case where $5$ isn't already used as the last digit.
Execute — Answer: D
5.OA.B.3 Step 1 - Pin down the digit pool and the cases.
- Since $5$ is the largest digit, every digit lies in $\{0, 1, 2, 3, 4, 5\}$.
- Divisibility by $5$ forces $D \in \{0, 5\}$, so the count splits into two non-overlapping cases.
💡 Splitting by the value of the last digit is a forced-cases move — exactly the Grade 5 habit of generating cases from a rule.
3.OA.A.1 Step 2 - Case 1 ($D = 5$).
- The digit $5$ is already used, so the $"5$ must appear$"$ rule is automatically satisfied.
- The other three slots $A, B, C$ get distinct digits from $\{0, 1, 2, 3, 4\}$.
- For the thousands slot, $A \neq 0$ leaves $\{1, 2, 3, 4\}$, giving $4$ choices.
- For $B$, any of the remaining $5 - 1 = 4$ digits in $\{0, 1, 2, 3, 4\}$ works (the $0$ is allowed here).
- For $C$, $3$ digits remain.
💡 Multiplying "choices per slot" is the Grade 3 idea of products as repeated grouping — slot-by-slot counting.
4.OA.A.3 Step 3 - Case 2 ($D = 0$).
- The digit $0$ is now used at the end, so $A, B, C$ are distinct digits drawn from $\{1, 2, 3, 4, 5\}$.
- The rule "$5$ must appear" is not yet satisfied (it isn't in slot $D$), so $5$ must occupy one of $A, B, C$.
- Pick the position of $5$ first ($3$ choices), then fill the other two positions from the remaining $\{1, 2, 3, 4\}$: $4$ choices, then $3$ choices.
💡 Anchoring on the forced digit $5$ and counting around it is a Grade 4 multi-step counting move.
4.OA.A.3 Step 4 - Add the two cases.
- Since they are disjoint (one has $D = 5$, the other $D = 0$), the total count is the sum.
💡 Disjoint-case sums match the Grade 4 multi-step whole-number reasoning pattern.
5.OA.B.3 Pin down the digit pool and the cases. Since $5$ is the largest digit, every dig 3.OA.A.1 Case 1 ($D = 5$). The digit $5$ is already used, so the $"5$ must appear$"$ rule 4.OA.A.3 Case 2 ($D = 0$). The digit $0$ is now used at the end, so $A, B, C$ are distinc 4.OA.A.3 Add the two cases. Since they are disjoint (one has $D = 5$, the other $D = 0$), Review
Reasonableness: Sanity-check the sizes. The total count of 4-digit numbers with distinct digits drawn from $\{0,1,2,3,4,5\}$ and $A \neq 0$ is $5 \times 5 \times 4 \times 3 = 300$. Of those, the fraction with $D \in \{0, 5\}$ should be roughly $\tfrac{2}{6} = \tfrac{1}{3}$, giving about $100$. Our $48 + 36 = 84$ sits just below that estimate — lower because requiring the digit $5$ to actually appear trims out the all-$\{0,1,2,3,4\}$ numbers. $84$ matches choice (D) and is small enough to be plausible, large enough to be more than the trivial counts in (A)-(B).
Alternative: Tool #16 (Count the Complement). In Case 2 ($D = 0$), count without the "$5$ must appear" rule first: $A, B, C$ are distinct from $\{1, 2, 3, 4, 5\}$, giving $5 \times 4 \times 3 = 60$ arrangements. Then subtract the ones with no $5$ at all, where $A, B, C$ come from $\{1, 2, 3, 4\}$: $4 \times 3 \times 2 = 24$. Case 2 count $= 60 - 24 = 36$. Combined with Case 1's $48$, the total is again $48 + 36 = 84$, confirming (D).
CCSS standards used (min grade 5)
3.OA.A.1Interpret products of whole numbers as totals from groups of equal size (Multiplying choices per digit-slot ($4 \times 4 \times 3$ in Case 1, $4 \times 3$ for the remaining slots in Case 2) — the fundamental counting principle as repeated grouping.)4.OA.A.3Solve multi-step word problems using the four operations with whole numbers (Counting Case 2 by first placing the forced digit $5$ ($3$ choices) and then multiplying by the slot count $4 \times 3$, and combining the two cases with $48 + 36 = 84$.)5.OA.B.3Generate two numerical patterns using two given rules and identify relationships (Splitting the count into the two forced cases $D = 5$ and $D = 0$ from the divisibility-by-$5$ rule, then handling each by its own rule for the digit $5$.)
⭐ This AMC 8 problem only needs the Grade 5 habit of splitting into cases from a rule plus slot-by-slot multiplication you learned in Grade 3!
⭐ This AMC 8 problem only needs the Grade 5 habit of splitting into cases from a rule plus slot-by-slot multiplication you learned in Grade 3!
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