AMC 10 · 2023 · #7
Grade 4 arithmeticPick an answer.
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.
Count the year's digits
The year uses zero and three oddly.
Sort the year's digits and count each — Grade 2 odd/even classification.
2.OA.C.3Organize Information In More WaysCondition on the last four
Write the condition on the last four.
Odd + odd = even; the four extra digits must "fix" each odd column. Parity arithmetic, Grade 2.
Odd plus odd is even, so the extra digits have to fix each odd column.
▸ Why?
Two numbers of the same parity add to an even total, while a mismatched pair always lands on odd.
▸ Why?
Each digit's total count is the fixed part plus the free part, so the free part is forced.
List the possible multisets
List the possible digit multisets.
List the possibilities by case — Grade 4 "generate by a rule" enumeration.
4.OA.C.5Make A Systematic ListKill the impossible ones
Kill the ones that cannot be dates.
Real calendar rules — months 01–12, no day 00 or ＞ 31 — kill these multisets. Grade 3 multi-step reasoning.
3.OA.D.8Identify SubproblemsFind the candidate digit
Find which digits can appear.
Filter d by the constraint that some arrangement gives MM ≤ 12 — Grade 4 factor-style case checking.
4.OA.B.4Identify SubproblemsCount the first candidate
Count the dates for the first candidate.
Systematic listing of arrangements that pass the month/day filter — Grade 4 multi-step word problem.
4.OA.A.3Make A Systematic ListCount the second candidate
Count the second candidate's 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 them up
Adding gives 9.
Two case totals to add — Grade 2 fluency within 100.
2.NBT.B.5Identify SubproblemsThis AMC 12 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.