AMC 10 · 2006 · #21
Easy mode Grade 4A four-digit number runs from 1000 to 9999. How many of them have at least one digit that is a 2 or a 3?
Pick an answer.
AMC 10 2006 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: Count the four-digit positive integers (from $1000$ to $9999$) that use the digit $2$ or the digit $3$ at least once, anywhere in the number.
Givens: The numbers are four-digit positive integers, so they run from $1000$ to $9999$; A number qualifies if any one of its four digits is a $2$ or a $3$; Answer choices: (A) $2439$, (B) $4096$, (C) $4903$, (D) $4904$, (E) $5416$
Unknowns: How many four-digit integers contain at least one digit equal to $2$ or $3$
Understand
Restated: Count the four-digit positive integers (from $1000$ to $9999$) that use the digit $2$ or the digit $3$ at least once, anywhere in the number.
Givens: The numbers are four-digit positive integers, so they run from $1000$ to $9999$; A number qualifies if any one of its four digits is a $2$ or a $3$; Answer choices: (A) $2439$, (B) $4096$, (C) $4903$, (D) $4904$, (E) $5416$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The phrase "at least one" is the classic signal for Tool #16 (Count the Complement): counting numbers that contain a $2$ or $3$ directly forces messy overlapping cases, but the opposite group — numbers with NO $2$ and NO $3$ — is a single clean count. So the plan is total minus none. Tool #7 (Identify Subproblems) handles each of those two counts by picking one digit position at a time and multiplying the choices. Tool #3 (Eliminate Possibilities) gives a fast sanity check: more than half of all four-digit numbers should qualify, which already points to the largest choices.
Execute — Answer: E
4.OA.A.3 Step 1 Flip to the complement
- Counting "has a $2$ or $3$" head-on means juggling overlaps (a number might have a $2$ and a $3$, or two $2$s, and so on).
- Flip the question instead: count every four-digit number, then subtract the ones that avoid both $2$ and $3$ completely.
- What is left must contain at least one $2$ or $3$.
💡 Everything either has what you want or avoids it, so "the rest" is just the whole minus the avoiders.
4.NBT.B.5 Step 2 Count all four-digit numbers
- Build a four-digit number one place at a time.
- The first digit can be $1$ through $9$ ($9$ choices, since it cannot be $0$), and each of the other three digits can be $0$ through $9$ ($10$ choices each).
- Multiplying the independent choices gives the total.
💡 Choices made one slot at a time multiply, because every first digit pairs with every ending.
4.NBT.B.5 Step 3 Count numbers with no 2 and no 3
- Now forbid both $2$ and $3$ in every slot.
- The first digit must avoid $0$, $2$, and $3$, leaving $\{1,4,5,6,7,8,9\}$ — that is $7$ choices.
- Each of the other three digits must avoid only $2$ and $3$, leaving $\{0,1,4,5,6,7,8,9\}$ — that is $8$ choices each.
- Multiply: $7 \times 8 \times 8 \times 8 = 7 \times 512 = 3584$.
💡 Removing two forbidden digits shrinks each slot's menu, and the shrunk menus multiply the same way.
4.NBT.B.4 Step 4 Subtract to finish
- Take away the avoiders from the whole: $9000 - 3584 = 5416$.
- Those $5416$ numbers are exactly the ones that could not dodge both digits, so each contains at least one $2$ or $3$.
- That matches choice (E).
💡 The leftover after removing the avoiders is precisely the group you were counting.
4.OA.A.3 Counting "has a $2$ or $3$" head-on means juggling overlaps (a number might have 4.NBT.B.5 Build a four-digit number one place at a time. The first digit can be $1$ throug 4.NBT.B.5 Now forbid both $2$ and $3$ in every slot. The first digit must avoid $0$, $2$, 4.NBT.B.4 Take away the avoiders from the whole: $9000 - 3584 = 5416$. Those $5416$ number Review
Reasonableness: The answer $5416$ is well over half of the $9000$ total, which fits intuition: with four digit slots each having a fair chance of landing on a $2$ or $3$, it is likely that at least one does — so a majority qualifying is expected. This immediately kills (A) $2439$ (less than half) and makes the top choices credible. The avoider count checks out too: $8^3 = 512$ and $7 \times 512 = 3584$, and $3584 + 5416 = 9000$ restores the full total, so no numbers were lost or double-counted.
Alternative: One could try to count directly with inclusion-exclusion on "contains a $2$" and "contains a $3$," but that path is longer and error-prone. A cleaner confirmation: the fraction of four-digit numbers with no $2$ or $3$ is $\frac{7}{9} \times \left(\frac{8}{10}\right)^3 = \frac{7}{9} \times \frac{512}{1000}$, and applying that fraction to $9000$ again gives $3584$ avoiders, hence $5416$ that qualify — the same (E).
CCSS standards used (min grade 4)
4.OA.A.3Solve multistep word problems using the four operations (Framing the count as total minus the complement (numbers with no 2 or 3), a two-operation plan.)4.NBT.B.5Multiply multi-digit whole numbers using place-value strategies (Multiplying the independent digit choices: $9 \times 10^3 = 9000$ and $7 \times 8^3 = 3584$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Subtracting the avoiders from the total: $9000 - 3584 = 5416$.)
⭐ When a problem says "at least one," count everything and subtract the cases that have none — the leftover is your answer.
⭐ When a problem says "at least one," count everything and subtract the cases that have none — the leftover is your answer.
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