AMC 10 · 2023 · #9
Grade 4 arithmeticProblem
A digital display shows the current date as an 8-digit integer consisting of a 4-digit year, followed by a 2-digit month, followed by a 2-digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in 2023 will each digit appear an even number of times in the 8-digital display for that date?
Pick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: In the $8$-digit display $\text{YYYYMMDD}$, 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.
Givens: Year digits: $2, 0, 2, 3$ — the digit $2$ appears twice, $0$ once, $3$ once, all others zero times; Month $\text{MM}$ is between $01$ and $12$; Day $\text{DD}$ is a valid day for that month in $2023$ (not a leap year, so February has $28$ days); Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $8$, (E) $9$
Unknowns: The number of valid dates $\text{MMDD}$ such that the digit count of $\text{YYYYMMDD}$ is even for every digit
Understand
Restated: In the $8$-digit display $\text{YYYYMMDD}$, 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.
Givens: Year digits: $2, 0, 2, 3$ — the digit $2$ appears twice, $0$ once, $3$ once, all others zero times; Month $\text{MM}$ is between $01$ and $12$; Day $\text{DD}$ is a valid day for that month in $2023$ (not a leap year, so February has $28$ days); Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $8$, (E) $9$
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #2 Make a Systematic List, #7 Identify Subproblems
The first move is Tool #15 (Reorganize) — drop the calendar story and re-frame the question as a digit-parity bookkeeping problem on $\text{MMDD}$. The year $2023$ already has odd counts only at digits $0$ and $3$, so the four digits of $\text{MMDD}$ 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 $\text{MM}/\text{DD}$.
Execute — Answer: E
2.OA.C.3 Step 1 - Count the digits of the year $2023$.
- The multiset is $\{2, 0, 2, 3\}$, so the digit $2$ appears twice (even), $0$ once (odd), $3$ once (odd), and every other digit zero times (even).
💡 Sort the year's digits and count each — Grade 2 odd/even classification.
2.OA.C.3 Step 2 - For the full $8$-digit string to have every digit appearing an even number of times, the four $\text{MMDD}$ digits must flip the year's parities.
- So the $\text{MMDD}$ digit multiset must contain an odd number of $0$'s, an odd number of $3$'s, and an even number of every other digit.
💡 Odd + odd = even; the four extra digits must "fix" each odd column. Parity arithmetic, Grade 2.
4.OA.C.5 Step 3 - List the digit multisets compatible with the parity rule on exactly four slots.
- The odd counts of $0$'s and $3$'s alone use $1+1 = 2$, $1+3 = 4$, or $3+1 = 4$ slots; any extra slots must come in even pairs of some other digit $d$.
- The valid multisets are $\{0, 3, d, d\}$ for some digit $d$, $\{0, 3, 3, 3\}$, and $\{0, 0, 0, 3\}$.
💡 List the possibilities by case — Grade 4 "generate by a rule" enumeration.
3.OA.D.8 Step 4 - Subproblem A — multiset $\{0, 3, 3, 3\}$ and $\{0, 0, 0, 3\}$.
- Any arrangement into $\text{MM} \text{DD}$ either gives $\text{MM} \ge 13$ or $\text{DD}$ with a leading $3$ ($\ge 30$ in a month with three $3$'s or $0$'s), or has $\text{MM} = 00$ or $\text{DD} = 00$.
- Check the candidates: from $\{0,3,3,3\}$ the only $\text{MM}$ values $\le 12$ are $03$ and $30$ — but $30 > 12$ — leaving $\text{MM}=03$ and $\text{DD}=33$, invalid.
- From $\{0,0,0,3\}$ valid $\text{MM}$ would be $00$ or $03$ or $30$; only $03$ qualifies, giving $\text{DD}=00$, invalid.
- Zero dates from these two multisets.
💡 Real calendar rules — months $01$–$12$, no day $00$ or $> 31$ — kill these multisets. Grade 3 multi-step reasoning.
4.OA.B.4 Step 5 - Subproblem B — multiset $\{0, 3, d, d\}$ with $d \ne 0, 3$.
- For $\text{MM}$ to be valid ($01$–$12$), the digit $d$ must let $\text{MM}$ stay $\le 12$.
- Check $d$ by case: $d = 1$ works (gives months $01$, $10$, $11$, $03$); $d = 2$ works ($\text{MM} = 02, 03, 12$ possible); $d \ge 4$ fails because $\text{MM}$ would need to use two large digits and still be $\le 12$, but the two $d$'s can only form $\text{MM}=dd \ge 44$ unless one of $\{0, 3\}$ takes the lead — and then $\text{DD}$ would need digits $\ge 30$ that exceed any month's day count.
💡 Filter $d$ by the constraint that some arrangement gives $\text{MM} \le 12$ — Grade 4 factor-style case checking.
4.OA.A.3 Step 6 - Sub-list $d = 1$: digit multiset $\{0, 1, 1, 3\}$.
- Pick $\text{MM}$ from valid month-strings using two of the four digits, then $\text{DD}$ from the remaining two.
- Valid $(\text{MM}, \text{DD})$ pairs that also form a real calendar day: $(01, 13), (01, 31), (03, 11), (10, 13), (10, 31), (11, 03), (11, 30)$.
- Seven dates.
💡 Systematic listing of arrangements that pass the month/day filter — Grade 4 multi-step word problem.
4.OA.A.3 Step 7 - Sub-list $d = 2$: digit multiset $\{0, 2, 2, 3\}$.
- Try every valid month: $\text{MM} = 02$ leaves $\{2, 3\}$ for $\text{DD}$ giving $23$ or $32$ — only $23$ is valid (February has $28$ days in $2023$).
- $\text{MM} = 03$ leaves $\{2, 2\}$ giving $\text{DD} = 22$, valid.
- $\text{MM} = 12$ leaves $\{0, 3\}$ giving $03$ or $30$ — but the multiset requires two $2$'s, and using $12$ as month consumes only one $2$, leaving one $2$ unused; actually $\text{MM} = 12$ doesn't fit the multiset $\{0,2,2,3\}$ at all since it would use digits $1, 2$ and we have no $1$.
- Final count: $(02, 23)$ and $(03, 22)$.
- Two dates.
💡 Same enumeration discipline as $d=1$, with the calendar killing the $32$ candidate. Grade 4 multi-step.
2.NBT.B.5 Step 8 - Combine.
- Multiset $\{0,1,1,3\}$ gives $7$ dates, $\{0,2,2,3\}$ gives $2$ dates, and the other multisets give $0$.
- Total $7 + 2 = 9$, matching choice (E).
💡 Two case totals to add — Grade 2 fluency within $100$.
2.OA.C.3 Count the digits of the year $2023$. The multiset is $\{2, 0, 2, 3\}$, so the di 2.OA.C.3 For the full $8$-digit string to have every digit appearing an even number of ti 4.OA.C.5 List the digit multisets compatible with the parity rule on exactly four slots. 3.OA.D.8 Subproblem A — multiset $\{0, 3, 3, 3\}$ and $\{0, 0, 0, 3\}$. Any arrangement i 4.OA.B.4 Subproblem B — multiset $\{0, 3, d, d\}$ with $d \ne 0, 3$. For $\text{MM}$ to b 4.OA.A.3 Sub-list $d = 1$: digit multiset $\{0, 1, 1, 3\}$. Pick $\text{MM}$ from valid m 4.OA.A.3 Sub-list $d = 2$: digit multiset $\{0, 2, 2, 3\}$. Try every valid month: $\text 2.NBT.B.5 Combine. Multiset $\{0,1,1,3\}$ gives $7$ dates, $\{0,2,2,3\}$ gives $2$ dates, Review
Reasonableness: Spot-check each of the $9$ dates by concatenating with $2023$ and tallying digits: e.g. $20230113 \to$ 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 ✓. $20230223 \to \{2,0,2,3,0,2,2,3\}$: $0$ 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.
Alternative: Tool #6 (Guess & Check) — direct calendar scan: walk through each of the $12$ months, and for each month enumerate the valid days, concatenate with $2023$ and tally digit counts on the fly. Slower (you check $\sim 365$ candidates instead of $\sim 20$ multiset arrangements), but it requires no parity reasoning — only patience and a list. The answer comes out the same: $9$.
CCSS standards used (min grade 4)
2.OA.C.3Determine whether a group of objects has an odd or even number (Tallying parities of each digit in $2023$ ($0$: odd, $3$: odd, $2$: even, rest: even) and propagating that parity bookkeeping to $\text{MMDD}$.)4.OA.C.5Generate a number or shape pattern following a given rule (Enumerating the digit-multisets on $4$ slots compatible with the parity rule — $\{0,3,d,d\}$, $\{0,3,3,3\}$, $\{0,0,0,3\}$.)4.OA.B.4Find all factor pairs / consider all possibilities for a whole number (Filtering the family $\{0, 3, d, d\}$ on the value of $d$ so that some arrangement yields a valid month $\text{MM} \le 12$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Inside each multiset, building all valid $(\text{MM}, \text{DD})$ assignments while honoring calendar constraints (month $\le 12$, day $\le$ days-in-month).)3.OA.D.8Solve two-step word problems using the four operations (Showing that multisets $\{0,3,3,3\}$ and $\{0,0,0,3\}$ produce no valid calendar date by checking each possible $\text{MM}/\text{DD}$ split.)2.NBT.B.5Fluently add and subtract within 100 (Adding the case totals $7 + 2 = 9$ at the end.)
⭐ 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 $\text{MMDD}$ digits, then list the calendar dates that fit.
⭐ 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 $\text{MMDD}$ digits, then list the calendar dates that fit.
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