AMC 8 · 2002 · #4
Grade 4 number-theoryProblem
The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?
Pick an answer.
AMC 8 2002 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: The year $2002$ reads the same forward and backward, so it is a palindrome. Find the next year after $2002$ that is also a palindrome, then multiply its four digits together.
Givens: A palindrome reads the same left-to-right and right-to-left; $2002$ is a $4$-digit palindromic year; We want the smallest palindromic year greater than $2002$; Answer choices: (A) $0$, (B) $4$, (C) $9$, (D) $16$, (E) $25$
Unknowns: The product of the four digits of the next palindromic year
Understand
Restated: The year $2002$ reads the same forward and backward, so it is a palindrome. Find the next year after $2002$ that is also a palindrome, then multiply its four digits together.
Givens: A palindrome reads the same left-to-right and right-to-left; $2002$ is a $4$-digit palindromic year; We want the smallest palindromic year greater than $2002$; Answer choices: (A) $0$, (B) $4$, (C) $9$, (D) $16$, (E) $25$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #6 Guess and Check
A $4$-digit palindrome always has the shape ABBA — the thousands digit matches the units digit, and the hundreds digit matches the tens digit. Tool #5 (Look for a Pattern) captures that shape. Once we lock the shape in, Tool #6 (Guess and Check) finishes the job: keep A $= 2$ to stay close to $2002$, and try the next value of B.
Execute — Answer: B
4.NBT.A.2 Step 1 - Write down the shape of a $4$-digit palindrome.
- Call the thousands digit A and the hundreds digit B.
- Then the tens digit must also be B and the units digit must also be A, so the year reads ABBA.
💡 Each digit gets a place value; the palindrome rule pins the units digit to the thousands digit and the tens digit to the hundreds digit.
4.NBT.A.2 Step 2 - Match $2002$ to the shape.
- The thousands digit is $2$, so A $= 2$.
- The hundreds digit is $0$, so B $= 0$.
- The tens digit ($0$) and units digit ($2$) confirm the pattern.
💡 Reading $2002$ as ABBA shows the starting values we need to bump.
4.NBT.A.2 Step 3 - To stay as close to $2002$ as possible, keep A $= 2$ and try the next value of B.
- If we change A to $3$ the smallest palindrome jumps to $3003$ — far too big.
- So increase B from $0$ to $1$ and check the result.
💡 Increasing B by $1$ raises the year by $110$ (B in the hundreds and B in the tens), which is the smallest possible jump while A stays at $2$.
3.OA.C.7 Step 4 Multiply the four digits of $2112$.
💡 Multiplying by $1$ leaves the value unchanged, so the product is just $2 \times 2$.
4.NBT.A.2 Write down the shape of a $4$-digit palindrome. Call the thousands digit A and t 4.NBT.A.2 Match $2002$ to the shape. The thousands digit is $2$, so A $= 2$. The hundreds 4.NBT.A.2 To stay as close to $2002$ as possible, keep A $= 2$ and try the next value of B 3.OA.C.7 Multiply the four digits of $2112$. Review
Reasonableness: Verify $2112$ really is a palindrome: read it backward and you get $2112$ — same number. Also check that no palindrome sits between $2002$ and $2112$. Any palindrome starting with $2$ has the form $\overline{2BB2}$, and the only choices for B between $0$ and $1$ are exactly those two values, giving $2002$ and $2112$. Nothing in between. Finally, the product $2 \times 1 \times 1 \times 2 = 4$ matches choice (B), and the trap answers $0, 9, 16, 25$ correspond to mistakenly using $2020$, $2332$, $2222$, or $2552$ — none of which is the next palindrome after $2002$.
Alternative: Tool #2 (Make a Systematic List): write down each year from $2003$ upward and reverse its digits. $2003 \to 3002$ (no), $2004 \to 4002$ (no), ..., $2112 \to 2112$ (yes). The first match is $2112$, and its digits multiply to $4$. Slower than the pattern method, but it forces you to check every candidate, leaving no doubt.
CCSS standards used (min grade 4)
4.NBT.A.2Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form (Recognizing the place value of each digit in a $4$-digit year and matching it to the ABBA palindrome shape to identify $2112$ as the next palindromic year.)3.OA.C.7Fluently multiply and divide within 100 (Computing the product of the four digits $2 \times 1 \times 1 \times 2 = 4$.)
⭐ A $4$-digit palindrome always reads ABBA, so the next one after $2002$ just needs the smallest bump in the middle digit — that gives $2112$, and its digits multiply to $4$.
⭐ A $4$-digit palindrome always reads ABBA, so the next one after $2002$ just needs the smallest bump in the middle digit — that gives $2112$, and its digits multiply to $4$.
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