AMC 10 · 2023 · #9
Easy mode Grade 4Problem
Imagine a digital display showing the date as an 8-digit number. The first 4 digits are the year, the next 2 are the month, and the last 2 are the day.
For example, April 28, 2023 shows up as 20230428.
Now look at all the dates in the year 2023. For some of these dates, every digit from 0 to 9 appears an even number of times in the 8-digit display. (Digits that don't show up at all count as appearing 0 times, which is even.)
How many dates in 2023 have this property?
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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