AMC 10 · 2016 · #6

Grade 4 counting
Archetype Casework by Position / Vertex Type └ Sub-archetype Digit Place Position
place-valuedigit-counting systematic-enumeration ↑ Prerequisites: place-value
📏 Long solution 💡 3 insights
📘 View easy version →
Problem
Ximena writes the whole numbers 1 through 30. Emilio copies them but turns every digit 2 into a digit 1. Find how much bigger Ximena's total is than Emilio's total.

Pick an answer.

(A)
13
(B)
26
(C)
102
(D)
103
(E)
110

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Adding all thirty numbers twice is slow and easy to slip on. Tool #16 (Change Focus) drops both full sums and looks only at the gap between them: a number changes its value only where a digit 2 becomes a digit 1, so the whole difference is built from those single-digit drops. Tool #7 (Identify Subproblems) then splits those drops by place value — tens-place 2s cost 10 each, ones-place 2s cost 1 each — because the two cases shrink the number by different amounts. Tool #2 (Make a Systematic List) just counts how many 2s sit in each place across 1 to 30.

1STEP 1

Track only the drops

Emilio's numbers differ only where a 2 became a 1, so the gap between the sums is just the total lost in those swaps.

Ximena - Emilio = Σ (value lost at each digit 2)
2STEP 2

Split by place value

A tens-place 2 drops from 20 to 10 (a loss of 10); a ones-place 2 drops from 2 to 1 (a loss of 1), so handle the two places separately.

tens 2 → 1: -10 ones 2 → 1: -1
3STEP 3

Count the tens-place 2s

The tens digit is 2 for exactly the ten numbers 20–29, each losing 10, for a total of 10 × 10 = 100.

10 numbers × 10 = 100
4STEP 4

Count the ones-place 2s

Only 2, 12, and 22 end in a 2; each loses 1, for a total of 3 × 1 = 3 (22's tens 2 was already counted above).

3 numbers × 1 = 3
5STEP 5

Add the two drops

Add the two losses: 100 + 3 = 103, so Ximena's sum tops Emilio's by 103 — choice (D).

100 + 3 = 103 → (D)
Answer
103
The tens-place loss must be the dominant piece, and 100 from ten numbers each dropping 10 is exactly that, so the answer should be just above 100 — and 103 is. Spot-check the ones: 22 becomes 11, a drop of 11 = 10 + 1, which correctly counts once in the tens total and once in the ones total. Choices 13 and 26 are far too small (they ignore the ten numbers in the twenties), and 110 would wrongly assume ten ones-place 2s, so 103 is the only sensible value.
💡Key takeaway

Only the digit 2s change, so just add up what each one loses — 10 for every tens-place 2 and 1 for every ones-place 2.

  • Track only the drops
  • Split by place value
  • Count the tens-place 2s
  • Count the ones-place 2s
  • Add the two drops

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