AMC 10 · 2022 · #3
Easy mode Grade 4Problem
Think about all the three-digit positive integers, from 100 all the way up to 999.
Each one has three digits. Some of those digits are even (0, 2, 4, 6, 8) and some are odd (1, 3, 5, 7, 9).
For each number, count how many of its digits are even. We only want the numbers where that count is an odd number — meaning the number has exactly one even digit, or exactly three even digits.
How many three-digit numbers are like that?
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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