AMC 10 · 2022 · #3
Grade 4 arithmeticProblem
How many three-digit positive integers have an odd number of even digits?
Pick an answer.
AMC 10 2022 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: Among all three-digit numbers (from $100$ to $999$), how many have an odd number of even digits? Each digit is independently classified as even ($\{0, 2, 4, 6, 8\}$) or odd ($\{1, 3, 5, 7, 9\}$). "Odd number of even digits" means exactly $1$ or exactly $3$ even digits.
Givens: Range: three-digit positive integers, $100 \le n \le 999$; Even-digit set $E = \{0, 2, 4, 6, 8\}$, $|E| = 5$; Odd-digit set $O = \{1, 3, 5, 7, 9\}$, $|O| = 5$; First digit (hundreds place) cannot be $0$; "Odd number of even digits" = exactly $1$ even digit OR exactly $3$ even digits; Answer choices: (A) $150$, (B) $250$, (C) $350$, (D) $450$, (E) $550$
Unknowns: Count of three-digit integers with an odd number of even digits
Understand
Restated: Among all three-digit numbers (from $100$ to $999$), how many have an odd number of even digits? Each digit is independently classified as even ($\{0, 2, 4, 6, 8\}$) or odd ($\{1, 3, 5, 7, 9\}$). "Odd number of even digits" means exactly $1$ or exactly $3$ even digits.
Givens: Range: three-digit positive integers, $100 \le n \le 999$; Even-digit set $E = \{0, 2, 4, 6, 8\}$, $|E| = 5$; Odd-digit set $O = \{1, 3, 5, 7, 9\}$, $|O| = 5$; First digit (hundreds place) cannot be $0$; "Odd number of even digits" = exactly $1$ even digit OR exactly $3$ even digits; Answer choices: (A) $150$, (B) $250$, (C) $350$, (D) $450$, (E) $550$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
"How many three-digit numbers with __" is a counting problem. Tool #2 (Systematic List) gives us the order: list every digit-pattern by where the even digits sit. The valid patterns split into Tool #7 (Subproblems) — Case A: exactly one even digit (patterns EOO, OEO, OOE); Case B: all three even (EEE). For each pattern we use the multiplication principle, remembering the hundreds digit can't be $0$. Tool #3 (Eliminate) gives a sanity range — the answer must be under $900$ and one of the five listed values.
Execute — Answer: D
4.OA.A.3 Step 1 - Set up the digit menus.
- The hundreds digit (H) has $9$ choices ($1$-$9$); the tens digit (T) and ones digit (U) each have $10$ choices ($0$-$9$).
- Split each menu by parity: H-even $\in \{2,4,6,8\}$ (4 choices), H-odd $\in \{1,3,5,7,9\}$ (5 choices).
- T and U each have $5$ even and $5$ odd.
💡 Listing how many choices each slot has under each parity is Grade 4 "multi-step word problem" bookkeeping.
3.OA.A.3 Step 2 - Case A — exactly one even digit.
- List the three patterns and multiply slot counts.
- Pattern EOO: $\text{H-even} \cdot \text{T-odd} \cdot \text{U-odd} = 4 \cdot 5 \cdot 5 = 100$.
- Pattern OEO: $5 \cdot 5 \cdot 5 = 125$.
- Pattern OOE: $5 \cdot 5 \cdot 5 = 125$.
💡 Multiply the number of choices for each slot — Grade 3 "multiplication word problems". The hundreds slot loses the digit $0$, so its even count drops from $5$ to $4$.
3.NBT.A.2 Step 3 Add the Case A patterns to get the one-even-digit subtotal.
💡 Add three subtotals — Grade 3 "fluently add within $1000$".
3.OA.A.3 Step 4 - Case B — all three digits even (pattern EEE).
- Hundreds-even has $4$ choices, tens-even and ones-even each have $5$.
💡 Same multiplication-principle drill; only the hundreds slot is restricted.
4.OA.A.3 Step 5 Combine Case A and Case B for the final count.
💡 Sum the two disjoint cases — Grade 4 "multi-step word problem with four operations".
4.OA.A.3 Set up the digit menus. The hundreds digit (H) has $9$ choices ($1$-$9$); the te 3.OA.A.3 Case A — exactly one even digit. List the three patterns and multiply slot count 3.NBT.A.2 Add the Case A patterns to get the one-even-digit subtotal. 3.OA.A.3 Case B — all three digits even (pattern EEE). Hundreds-even has $4$ choices, ten 4.OA.A.3 Combine Case A and Case B for the final count. Review
Reasonableness: Symmetry check. The total number of three-digit integers is $900$. For each three-digit number the parity count of even digits is $0, 1, 2,$ or $3$. Pair up complementary cases: numbers with $0$ even digits ($\text{OOO} = 5 \cdot 5 \cdot 5 = 125$) plus numbers with $2$ even digits ($\text{EEO} + \text{EOE} + \text{OEE} = 100 + 100 + 125 = 325$) gives $450$ for the "even count of even digits" side. The remaining $900 - 450 = 450$ must be the "odd count of even digits" side — exactly our answer (D). The two sides split $900$ down the middle.
Alternative: Tool #16 (Change Focus / Complement). Count the complement: three-digit numbers with an EVEN number of even digits ($0$ or $2$). $0$-even is $\text{OOO} = 125$; $2$-even patterns EEO/EOE/OEE total $100 + 100 + 125 = 325$. Sum $= 450$. Subtract from total: $900 - 450 = 450$. Same answer (D) — and the matching counts hint at a deeper parity symmetry worth noting.
CCSS standards used (min grade 4)
3.NBT.A.2Fluently add and subtract within 1000 (Adding the case subtotals $100 + 125 + 125$ and $350 + 100$.)3.OA.A.3Solve multiplication and division word problems within 100 (Applying the multiplication principle to each three-slot digit pattern ($4 \times 5 \times 5$, $5 \times 5 \times 5$, etc.).)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Splitting the count into cases (1 even or 3 even), computing each, and combining the totals.)
⭐ This AMC 10 problem only needs Grade 4 "organize cases, multiply slot choices, then add" — exactly $1$ even digit gives $350$ numbers, exactly $3$ even gives $100$, and the total is $450$.
⭐ This AMC 10 problem only needs Grade 4 "organize cases, multiply slot choices, then add" — exactly $1$ even digit gives $350$ numbers, exactly $3$ even gives $100$, and the total is $450$.
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