AMC 8 · 2019 · #13
Easy mode Grade 4Problem
A palindrome is a number that reads the same forward and backward. For example, 12321 is a palindrome, and so is 77.
Picture all the two-digit numbers that are palindromes. (These are 11, 22, 33, and so on.)
We want to find a three-digit number N with two properties. First, N is the sum of three different two-digit palindromes. Second, N itself is not a palindrome.
Among all numbers that fit both rules, N is the smallest one.
What is the sum of the digits of N?
Pick an answer.
AMC 8 2019 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 palindrome reads the same forwards and backwards. Let $N$ be the SMALLEST three-digit integer that (i) is itself NOT a palindrome and (ii) can be written as the sum of three DISTINCT two-digit palindromes. Find the sum of the digits of $N$.
Givens: Definition: a palindrome reads the same left-to-right and right-to-left (e.g. $12321$); $N$ is a three-digit integer (so $100 \le N \le 999$); $N$ is the sum of three DISTINCT two-digit palindromes; $N$ itself is NOT a palindrome; We want the LEAST such $N$; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Unknowns: The value of $N$, and then the sum of its digits
Understand
Restated: A palindrome reads the same forwards and backwards. Let $N$ be the SMALLEST three-digit integer that (i) is itself NOT a palindrome and (ii) can be written as the sum of three DISTINCT two-digit palindromes. Find the sum of the digits of $N$.
Givens: Definition: a palindrome reads the same left-to-right and right-to-left (e.g. $12321$); $N$ is a three-digit integer (so $100 \le N \le 999$); $N$ is the sum of three DISTINCT two-digit palindromes; $N$ itself is NOT a palindrome; We want the LEAST such $N$; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #2 Make a Systematic List, #6 Guess and Check
The two-digit palindromes form a tiny pattern: $11, 22, 33, \ldots, 99$ — Tool #5 (Pattern) immediately spots that every one of them equals $11 \times k$ for $k = 1, \ldots, 9$. That single observation forces the sum $N$ to be a multiple of $11$, which collapses the search space from $900$ three-digit numbers down to a short list. Tool #2 (Systematic List) then walks the three-digit multiples of $11$ in order ($110, 121, 132, \ldots$) to find the smallest one that is NOT a palindrome. Finally Tool #6 (Guess and Check) confirms that the winning candidate $110$ can actually be written as a sum of three DISTINCT two-digit palindromes. We deliberately avoid Tool #13 (Algebra) — the pattern + listing path is much cleaner.
Execute — Answer: A
1.NBT.B.2 Step 1 - List every two-digit palindrome.
- A two-digit number $\overline{ab}$ is a palindrome exactly when its two digits are equal, so the list is $11, 22, 33, 44, 55, 66, 77, 88, 99$ — only nine numbers.
💡 Understanding that a two-digit number is "tens digit, then ones digit" is the Grade 1 place-value idea — it tells us exactly which two-digit numbers can be palindromes.
4.NBT.B.5 Step 2 - Spot the pattern.
- Every entry in the list is a multiple of $11$: $11 = 11 \cdot 1$, $22 = 11 \cdot 2$, $\ldots$, $99 = 11 \cdot 9$.
- (The reason: a two-digit palindrome $\overline{aa}$ equals $10a + a = 11a$.)
💡 Rewriting $10a + a$ as $11 \cdot a$ is the Grade 4 multi-digit multiplication / distributive-property move that turns the list into one clean pattern.
4.OA.B.4 Step 3 - Because $N = P_1 + P_2 + P_3$ where each $P_i$ is a multiple of $11$, the sum $N$ is also a multiple of $11$.
- So we may restrict the search to three-digit multiples of $11$.
💡 Adding multiples of $11$ gives another multiple of $11$ — a Grade 4 "multiples of a whole number" property that prunes the search.
2.NBT.A.1 Step 4 - Make a systematic list of three-digit multiples of $11$ in increasing order and test each one for the "not a palindrome" condition.
- $110$?
- Not a palindrome (reversed is $011$).
- $121$?
- IS a palindrome — reject.
- $132$?
- Not a palindrome — but it's already bigger than $110$.
💡 Reading the hundreds, tens, and ones digits of a three-digit number to check whether it reads the same backwards is exactly the Grade 2 place-value skill.
2.NBT.B.7 Step 5 - Verify that $110$ really can be made from three DISTINCT two-digit palindromes.
- Try the smallest two palindromes first: $11 + 22 = 33$, leaving $110 - 33 = 77$.
- And $77$ is itself a two-digit palindrome, distinct from $11$ and $22$.
- So $110 = 11 + 22 + 77$ — the construction works.
💡 Adding three two-digit numbers to hit a target within $1000$ is a Grade 2 addition strategy — exactly what guess-and-check needs here.
1.OA.A.2 Step 6 Compute the sum of the digits of $N = 110$.
💡 Adding three single digits is the Grade 1 "sum of three whole numbers within $20$" standard.
1.NBT.B.2 List every two-digit palindrome. A two-digit number $\overline{ab}$ is a palindr 4.NBT.B.5 Spot the pattern. Every entry in the list is a multiple of $11$: $11 = 11 \cdot 4.OA.B.4 Because $N = P_1 + P_2 + P_3$ where each $P_i$ is a multiple of $11$, the sum $N 2.NBT.A.1 Make a systematic list of three-digit multiples of $11$ in increasing order and 2.NBT.B.7 Verify that $110$ really can be made from three DISTINCT two-digit palindromes. 1.OA.A.2 Compute the sum of the digits of $N = 110$. Review
Reasonableness: The answer choices are all single digits ($2$ through $6$), which is consistent with $N$ being a small three-digit number — and $1 + 1 + 0 = 2$ matches choice (A). We should also confirm we didn't miss a smaller candidate: any $N < 110$ is at most two-digit, so $110$ is the smallest three-digit multiple of $11$ at all; and the only smaller three-digit non-multiple-of-$11$ values can't be sums of three two-digit palindromes (which are all multiples of $11$). So $N = 110$ is forced and the digit sum $2$ is correct.
Alternative: Tool #6 (Guess and Check) on the choices: the digit sum of a three-digit number is between $1$ and $27$, and choices (A)-(E) only run $2$ to $6$, so $N$ must have very small digits. The smallest three-digit number with digit sum in that range AND that is a multiple of $11$ AND not itself a palindrome is $110$ (digit sum $2$). Every choice except (A) would force a larger $N$ and is therefore not the LEAST. Eliminating (B)-(E) leaves (A).
CCSS standards used (min grade 4)
1.NBT.B.2Understand that the two digits of a two-digit number represent tens and ones (Recognizing which two-digit numbers are palindromes by checking that the tens digit equals the ones digit.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Rewriting a two-digit palindrome $\overline{aa}$ as $10a + a = 11 \cdot a$ to expose the multiple-of-$11$ structure.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Using the property that a sum of multiples of $11$ is itself a multiple of $11$, and listing three-digit multiples of $11$ ($110, 121, 132, \ldots$).)2.NBT.A.1Understand that the three digits of a three-digit number represent hundreds, tens, and ones (Reading the digits of $110, 121, 132, \ldots$ to check whether each three-digit candidate reads the same backwards.)2.NBT.B.7Add and subtract within 1000 using models or strategies (Verifying that $11 + 22 + 77 = 110$, the required sum of three distinct two-digit palindromes.)1.OA.A.2Solve word problems involving three whole numbers whose sum is within 20 (Computing the final digit sum $1 + 1 + 0 = 2$.)
⭐ This AMC 8 problem only needs Grade 4 "multiples of a whole number" you already know — once you see that every two-digit palindrome is a multiple of $11$, the whole puzzle solves itself!
⭐ This AMC 8 problem only needs Grade 4 "multiples of a whole number" you already know — once you see that every two-digit palindrome is a multiple of $11$, the whole puzzle solves itself!
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