AMC 10 · 2014 · #15

Grade 7 number-theorycounting
place-valuedigit-sumfundamental-counting-principlesymmetry-argument identify-subproblemssymmetry-argument ↑ Prerequisites: place-valuedigit-sum
📏 Long solution 💡 3 insights
Problem
Every five-digit palindrome is added together. Add the digits of that total.

Pick an answer.

(A)
9
(B)
18
(C)
27
(D)
36
(E)
45
How to solve
Strategy Organize Information in More Ways

Tool #15 (Reorganize) is the whole idea: adding 900 five-digit numbers one at a time is hopeless, but adding them slot by slot is three short computations. Tool #4 (Variable) gives the handle — write a palindrome as 10001a + 1010b + 100c so the slots become separate. Tool #7 (Subproblems) splits the job into "total of all a's", "total of all b's", "total of all c's". Tool #16 (Change Focus) supplies an independent second route: pair each palindrome with its digit-complement instead of summing slots, and every pair adds to the same amount.

1STEP 1

Count the palindromes

There are 900 such palindromes.

9 · 10 · 10 = 900 five-digit palindromes
2STEP 2

Expand by place value

Place value turns one into a weighted sum.

abcba = 10000a + 1000b + 100c + 10b + a = 10001a + 1010b + 100c
3STEP 3

Add by column, not by number

So the total adds up column by column.

S = 10001 Σ a + 1010 Σ b + 100 Σ c, each sum taken over all 900 triples
4STEP 4

Total each digit slot

Each digit slot totals easily.

Σ a = 45 · 100 = 4500, Σ b = 45 · 90 = 4050, Σ c = 45 · 90 = 4050
5STEP 5

Multiply out and add

Multiplying out gives a round total.

S = 10001 · 4500 + 1010 · 4050 + 100 · 4050 = 45,004,500 + 4,090,500 + 405,000 = 49,500,000
6STEP 6

Cross-check by pairing complements

Pairing complements confirms it.

abcba + a'b'c'b'a' = 100010 + 9090 + 900 = 110,000, so S = 450 · 110,000 = 49,500,000 ✓
7STEP 7

Add the digits of S

Its digits add to 18, choice (B).

4 + 9 + 5 + 0 + 0 + 0 + 0 + 0 = 18 → (B) 18
Answer
18
Divide back: 49,500,000/900 = 55,000, so the average five-digit palindrome is 55,000. That is exactly the midpoint of the smallest palindrome 10001 and the largest 99999, which is what the symmetry of the digit ranges predicts — a strong sign the total is right. The crude bounds agree too: 900 · 10001 = 9,000,900 ≤ S ≤ 900 · 99999 = 89,999,100. The step that is easy to get wrong is the multiplicity 90 for b and c: forgetting that a ≠ 0 and using 100 everywhere gives 4500 · (10001 + 1010 + 100) = 4500 · 11111 = 49,999,500, whose digit sum is 45 — which is exactly the trap answer (E) sitting in the choice list. The pairing route in the execute steps is immune to that slip, which is why it is worth running.
💡Key takeaway

Adding 900 numbers one at a time is hopeless, so add them column by column: each digit slot has a total you can get in one line. That gives S = 49,500,000, and its digits add to 18 — choice (B).

  • Count the palindromes
  • Expand by place value
  • Add by column, not by number
  • Total each digit slot
  • Multiply out and add
  • Cross-check by pairing complements
  • Add the digits of S