Competition · AMC preparation · step 4 of 4

AMC 8 · 2019 · #13

Grade 4 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration
multiplesdivisibility-rulesdigit-constraintssystematic-enumeration systematic-enumerationpattern-recognition ↑ Prerequisites: multiplesdivisibility-rules
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Problem
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.

Pick an answer.

(A)
2
(B)
3
(C)
4
(D)
5
(E)
6

AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

The two-digit palindromes form a tiny pattern: 11, 22, 33, …, 99 — Tool #5 (Pattern) immediately spots that every one of them equals 11 × k for k = 1, …, 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, …) 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.

1STEP 1

List the two-digit palindromes

List every two-digit palindrome — the two digits must match, so there are only nine of them.

{11, 22, 33, 44, 55, 66, 77, 88, 99}
2STEP 2

Spot the multiple of 11

Spot the pattern: each palindrome aa equals 10a + a = 11a, so all nine are multiples of 11.

aa = 10a + a = 11a
3STEP 3

Show the sum is a multiple

Since N sums three multiples of 11, N itself is a multiple of 11 — so search only three-digit multiples of 11.

N = 11k₁ + 11k₂ + 11k₃ = 11(k₁ + k₂ + k₃)
4STEP 4

Scan the three-digit multiples

Scan three-digit multiples of 11 upward: 110 is not a palindrome (121 is), so the least candidate is 110.

110, 121, 132, … ⟹ smallest non-palindrome = 110
5STEP 5

Check that 110 is buildable

Confirm 110 is buildable: 11 + 22 + 77 = 110 uses three distinct two-digit palindromes.

11 + 22 + 77 = 110
6STEP 6

Add the digits

Add the digits of N = 110: 1 + 1 + 0 = 2 → (A).

1 + 1 + 0 = 2 → (A)
Answer
2
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.
💡Key takeaway

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!

  • List the two-digit palindromes
  • Spot the multiple of 11
  • Show the sum is a multiple
  • Scan the three-digit multiples
  • Check that 110 is buildable
  • Add the digits

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