AMC 10 · 2023 · #9

Grade 4 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

digit-constraintsparitysystematic-enumerationplace-value caseworksystematic-enumerationidentify-subproblems ↑ Prerequisites: digit-constraintsparity

📏 Long solution 💡 3 insights

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Problem

In the 8-digit display YYYYMMDD, the year part is fixed as 2023. Count the dates within 2023 for which every digit from 0 to 9 appears an even number of times across the entire 8-digit string.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Organize Information in More Ways

The first move is Tool #15 (Reorganize) — drop the calendar story and re-frame the question as a digit-parity bookkeeping problem on MMDD. The year 2023 already has odd counts only at digits 0 and 3, so the four digits of MMDD must include an odd count of 0's, an odd count of 3's, and an even count of every other digit. With four slots that pins the digit multiset down to a tiny set of cases — perfect for Tool #2 (Systematic List). Tool #7 (Identify Subproblems) handles the calendar filter: for each candidate digit multiset, list the rearrangements that form a valid MM/DD.

1STEP 1

Counting the year 2023's digits {2, 0, 2, 3}, only 0 and 3 land on odd counts; 2 is even and every other digit is zero.

year parity: 0 → odd, 3 → odd, else → even

2STEP 2

So the four MMDD digits must flip those parities: an odd number of 0's and 3's, and an even count of every other digit.

#{0}_MMDD ∈ {1, 3}, #{3}_MMDD ∈ {1, 3}, #{d}_MMDD ∈ {0, 2, 4} for d ∉ {0, 3}

3STEP 3

On only four slots, that pins the digit multiset to {0, 3, d, d}, {0, 3, 3, 3}, or {0, 0, 0, 3}.

cases: {0, 3, d, d}, {0, 3, 3, 3}, {0, 0, 0, 3}

4STEP 4

Calendar rules (months 01–12, no day 00 or 33) kill {0, 3, 3, 3} and {0, 0, 0, 3}: 0 dates from both.

{0,3,3,3}: 0, {0,0,0,3}: 0

5STEP 5

For {0, 3, d, d} the month must stay ≤ 12, so only d = 1 and d = 2 can ever work; d ≥ 4 forces MM ≥ 44 or an impossible day.

candidate d: 1, 2

6STEP 6

For {0, 1, 1, 3}, the real calendar dates are 0113, 0131, 0311, 1013, 1031, 1103, 1130 — 7 dates.

d=1: {0113, 0131, 0311, 1013, 1031, 1103, 1130} → 7

7STEP 7

For {0, 2, 2, 3}, only 0223 (Feb 23) and 0322 (Mar 22) survive — 32 fails the calendar — giving 2 dates.

d=2: {0223, 0322} → 2

8STEP 8

Add the case totals: 7 + 2 = 9, which is choice (E).

7 + 2 = 9 → (E)

Answer

Spot-check each of the 9 dates by concatenating with 2023 and tallying digits: e.g. 20230113 → digits {2,0,2,3,0,1,1,3}: 0 twice, 1 twice, 2 twice, 3 twice — all even ✓. Similarly 20230131: {2,0,2,3,0,1,3,1} — 0 twice, 1 twice, 2 twice, 3 twice ✓. 20230223 → {2,0,2,3,0,2,2,3}: 0 twice, 2 four times, 3 twice ✓. The same parity verification works on the rest. Magnitude check: with 9 qualifying dates out of 365 days in 2023, the rate is about 2.5% — a believable order of magnitude for a fairly restrictive parity constraint.

💡Key takeaway

This AMC 10 problem only needs Grade 4 pattern enumeration you already know — make 2023's odd-count digits (0 and 3) come out even by carefully choosing the four MMDD digits, then list the calendar dates that fit.

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