AMC 8 · 2002 · #4
Easy mode Grade 4Problem
A palindrome is a number that reads the same forward and backward. For example, 2002 reads the same from left to right as it does from right to left, so 2002 is a palindrome.
Find the next year after 2002 that is also a palindrome. Then take the four digits of that year and multiply them all together.
What is that product?
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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