AMC 8 · 2019 · #13

Easy mode Grade 4

Problem

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 $N$ ?

$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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

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

#2 Make a Systematic List 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.

$$\{11,\;22,\;33,\;44,\;55,\;66,\;77,\;88,\;99\}$$

💡 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.

#5 Look for a Pattern 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 $10a + a = 11a$.)

$$\overline{aa} = 10a + a = 11a$$

💡 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.

#5 Look for a 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 $11$.

$$N = 11k_1 + 11k_2 + 11k_3 = 11(k_1 + k_2 + k_3)$$

💡 Adding multiples of $11$ gives another multiple of $11$ — a Grade 4 "multiples of a whole number" property that prunes the search.

#2 Make a Systematic List 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 $110$.

$$110,\;121,\;132,\;\ldots \quad\Longrightarrow\quad \text{smallest non-palindrome} = 110$$

💡 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.

#6 Guess and Check 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.

$$11 + 22 + 77 = 110$$

💡 Adding three two-digit numbers to hit a target within $1000$ is a Grade 2 addition strategy — exactly what guess-and-check needs here.

#5 Look for a Pattern 1.OA.A.2 Step 6

Compute the sum of the digits of $N = 110$.

$$1 + 1 + 0 = 2 \;\Rightarrow\; \textbf{(A)}$$

💡 Adding three single digits is the Grade 1 "sum of three whole numbers within $20$" standard.

[1] #2 1.NBT.B.2 List every two-digit palindrome. A two-digit number $\overline{ab}$ is a palindr

[2] #5 4.NBT.B.5 Spot the pattern. Every entry in the list is a multiple of $11$: $11 = 11 \cdot

[3] #5 4.OA.B.4 Because $N = P_1 + P_2 + P_3$ where each $P_i$ is a multiple of $11$, the sum $N

[4] #2 2.NBT.A.1 Make a systematic list of three-digit multiples of $11$ in increasing order and

[5] #6 2.NBT.B.7 Verify that $110$ really can be made from three DISTINCT two-digit palindromes.

[6] #5 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.2 Understand 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.5 Multiply 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.4 Find 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.1 Understand 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.7 Add 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.2 Solve 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 4)

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