Competition · AMC preparation · step 4 of 4
AMC 10 · 2023A · #9
Grade 4 arithmeticPick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Check the year's digit parities
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.
Sort the year's digits and count each — Grade 2 odd/even classification.
2.OA.C.3Organize Information In More WaysDecide what MMDD must supply
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.
Odd + odd = even; the four extra digits must "fix" each odd column. Parity arithmetic, Grade 2.
The extra digits must fix the parity of each column that the year left odd.
▸ Why?
Two numbers add to an even total exactly when they are both even or both odd.
▸ Why?
The full count of a digit is exactly the year's count plus the extra digits' count.
List the possible digit multisets
On only four slots, that pins the digit multiset to {0, 3, d, d}, {0, 3, 3, 3}, or {0, 0, 0, 3}.
List the possibilities by case — Grade 4 "generate by a rule" enumeration.
4.OA.C.5Make A Systematic ListRule out the two lopsided cases
Calendar rules (months 01–12, no day 00 or 33) kill {0, 3, 3, 3} and {0, 0, 0, 3}: 0 dates from both.
Real calendar rules — months 01–12, no day 00 or ＞ 31 — kill these multisets. Grade 3 multi-step reasoning.
3.OA.D.8Identify SubproblemsNarrow the repeated digit d
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.
Filter d by the constraint that some arrangement gives MM ≤ 12 — Grade 4 factor-style case checking.
4.OA.B.4Identify SubproblemsList the dates for d equals 1
For {0, 1, 1, 3}, the real calendar dates are 0113, 0131, 0311, 1013, 1031, 1103, 1130 — 7 dates.
Systematic listing of arrangements that pass the month/day filter — Grade 4 multi-step word problem.
4.OA.A.3Make A Systematic ListList the dates for d equals 2
For {0, 2, 2, 3}, only 0223 (Feb 23) and 0322 (Mar 22) survive — 32 fails the calendar — giving 2 dates.
Same enumeration discipline as d=1, with the calendar killing the 32 candidate. Grade 4 multi-step.
4.OA.A.3Make A Systematic ListAdd the two counts
Add the case totals: 7 + 2 = 9, which is choice (E).
Two case totals to add — Grade 2 fluency within 100.
2.NBT.B.5Identify SubproblemsThis 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.
- Check the year's digit parities
- Decide what MMDD must supply
- List the possible digit multisets
- Rule out the two lopsided cases
- Narrow the repeated digit d
- List the dates for d equals 1
- List the dates for d equals 2
- Add the two counts
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