AMC 8 · 2020 · #19
Grade 4 number-theorycountingProblem
A number is called flippy if its digits alternate between two distinct digits. For example, 2020 and 37373 are flippy, but 3883 and 123123 are not. How many five-digit flippy numbers are divisible by 15?
Pick an answer.
AMC 8 2020 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: A "flippy" number is one whose digits strictly alternate between two distinct digits — like $2020$ or $37373$, but not $3883$ (two $3$s in a row) or $123123$ (three different digits). How many five-digit flippy numbers are multiples of $15$?
Givens: The number has exactly $5$ digits, so the leading digit is not $0$; Digits alternate between two distinct digits — call them $a$ and $b$, with $a \ne b$; The number must be divisible by $15$; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $8$
Unknowns: The count of five-digit flippy numbers that are multiples of $15$
Understand
Restated: A "flippy" number is one whose digits strictly alternate between two distinct digits — like $2020$ or $37373$, but not $3883$ (two $3$s in a row) or $123123$ (three different digits). How many five-digit flippy numbers are multiples of $15$?
Givens: The number has exactly $5$ digits, so the leading digit is not $0$; Digits alternate between two distinct digits — call them $a$ and $b$, with $a \ne b$; The number must be divisible by $15$; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $8$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List
The condition "divisible by $15$" splits cleanly into two smaller, very different sub-conditions: divisible by $5$ (a last-digit rule) and divisible by $3$ (a digit-sum rule). Tool #7 (Identify Subproblems) lets us pin down one digit at a time — the $5$-rule forces the value of $a$, then the $3$-rule narrows down $b$. Once both digits are constrained, Tool #2 (Make a Systematic List) finishes the job: write the few remaining candidate numbers in order and count them. We deliberately avoid Tool #13 (Algebra) — divisibility rules and listing handle this with no equations needed.
Execute — Answer: B
4.NBT.A.2 Step 1 - Write the shape of a five-digit flippy number.
- Because the digits alternate between two distinct digits $a$ and $b$, the number must look like $ababa$ (positions $1, 3, 5$ are $a$; positions $2, 4$ are $b$), with $a \ne b$ and $a \ne 0$ since $a$ is the leading digit.
💡 Reading a multi-digit number by its place-value positions is exactly the Grade 4 "read and write multi-digit whole numbers" skill.
4.OA.B.4 Step 2 - Subproblem 1: divisibility by $5$.
- A whole number is a multiple of $5$ exactly when its last digit is $0$ or $5$.
- The last digit of $ababa$ is $a$.
- Since $a \ne 0$ (no leading-zero five-digit numbers), the only choice left is $a = 5$.
- So the number must be $5b5b5$ with $b \ne 5$.
💡 Recognizing multiples of $5$ by the last digit is part of the Grade 4 "factors and multiples" cluster.
4.OA.B.4 Step 3 - Subproblem 2: divisibility by $3$.
- A whole number is a multiple of $3$ exactly when its digit-sum is a multiple of $3$.
- Add the digits of $5b5b5$: three $5$s and two $b$s, giving $15 + 2b$.
- Since $15$ is already a multiple of $3$, we need $2b$ to be a multiple of $3$ too.
💡 The "sum-of-digits divisible by $3$" rule is a standard Grade 4 divisibility test that lives in the factors-and-multiples standard.
4.OA.B.4 Step 4 - Find the digits $b$ that make $2b$ a multiple of $3$.
- Since $2$ and $3$ share no common factor, $2b$ is a multiple of $3$ exactly when $b$ itself is.
- The single-digit multiples of $3$ are $0, 3, 6, 9$.
- Check each against the constraint $b \ne 5$: all four pass.
💡 Listing the one-digit multiples of $3$ is direct Grade 4 multiples reasoning.
4.OA.C.5 Step 5 - Systematic list of every valid number.
- Plug each allowed $b$ into $5b5b5$ in increasing order and count.
💡 Generating every number from a clear rule ("$5b5b5$ for each allowed $b$") and counting matches the Grade 4 "generate a pattern following a rule" standard.
4.NBT.A.2 Write the shape of a five-digit flippy number. Because the digits alternate betw 4.OA.B.4 Subproblem 1: divisibility by $5$. A whole number is a multiple of $5$ exactly w 4.OA.B.4 Subproblem 2: divisibility by $3$. A whole number is a multiple of $3$ exactly w 4.OA.B.4 Find the digits $b$ that make $2b$ a multiple of $3$. Since $2$ and $3$ share no 4.OA.C.5 Systematic list of every valid number. Plug each allowed $b$ into $5b5b5$ in inc Review
Reasonableness: Quick sanity check: $50505 \div 15 = 3367$, $53535 \div 15 = 3569$, $56565 \div 15 = 3771$, $59595 \div 15 = 3973$ — all whole numbers, so each really is a multiple of $15$. There is no fifth candidate to miss: any other $b \in \{0,\dots,9\}$ either fails the digit-sum test ($b \in \{1,2,4,7,8\}$) or equals $5$ (which would collapse the number to $55555$, violating $a \ne b$). The count $4$ matches answer (B), well within the answer-choice range $3$ to $8$.
Alternative: Tool #2 (Make a Systematic List) could be used from the very start: list all multiples of $15$ of the form $5b5b5$ by trying $b = 0, 1, 2, \dots, 9$ one by one and checking each against the divisibility-by-$3$ rule. That gives the same four numbers but does more arithmetic; using the divisibility rules first (Tool #7 subproblems) is faster.
CCSS standards used (min grade 4)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Writing the five-digit flippy number in the place-value form $ababa$ and recognizing that the leading digit cannot be $0$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Applying divisibility rules: multiples of $5$ end in $0$ or $5$; multiples of $3$ have digit-sum divisible by $3$; and listing single-digit multiples of $3$ as $\{0, 3, 6, 9\}$.)4.OA.C.5Generate a number or shape pattern following a given rule (Generating every valid $5b5b5$ number from the allowed values of $b$ and counting the resulting list of four numbers.)
⭐ This AMC 8 problem only needs Grade 4 divisibility rules — "ends in $0$ or $5$" for $5$, and "digits add up to a multiple of $3$" for $3$ — that you already know!
⭐ This AMC 8 problem only needs Grade 4 divisibility rules — "ends in $0$ or $5$" for $5$, and "digits add up to a multiple of $3$" for $3$ — that you already know!
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