AMC 8 · 2009 · #22
Easy mode Grade 5Problem
Look at the whole numbers from 1 up to 1000. That includes 1,2,3,…,999,1000.
Some of those numbers have the digit 1 somewhere in them — like 1, 13, 312, or 1000. Others have no 1 at all — like 2, 25, or 407.
How many of the numbers from 1 to 1000 have no digit 1 in them?
Pick an answer.
AMC 8 2009 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 whole numbers strictly between $1$ and $1000$ whose decimal representation contains no digit "$1$". So we need to count integers $n$ with $1 < n < 1000$ such that none of the digits of $n$ is $1$.
Givens: Range: whole numbers strictly between $1$ and $1000$ (so $n \in \{2, 3, \ldots, 999\}$); Forbidden digit: $1$ may not appear in any place value of $n$; Answer choices: (A) $512$, (B) $648$, (C) $720$, (D) $728$, (E) $800$
Unknowns: The number of integers in the range that contain no digit $1$
Understand
Restated: Count the whole numbers strictly between $1$ and $1000$ whose decimal representation contains no digit "$1$". So we need to count integers $n$ with $1 < n < 1000$ such that none of the digits of $n$ is $1$.
Givens: Range: whole numbers strictly between $1$ and $1000$ (so $n \in \{2, 3, \ldots, 999\}$); Forbidden digit: $1$ may not appear in any place value of $n$; Answer choices: (A) $512$, (B) $648$, (C) $720$, (D) $728$, (E) $800$
Plan
Primary tool: #13 Count the Possibilities
Secondary: #7 Identify Subproblems
We are counting how many length-$3$ digit strings avoid one specific digit, which is exactly the multiplication principle situation Tool #13 (Count the Possibilities) handles. The clean trick is to pad every number up to three digits (write $7$ as "$007$", $42$ as "$042$", etc.) — then every integer from $0$ to $999$ becomes a $3$-slot string, each slot independently picks from $\{0,1,2,\ldots,9\}$, and we just forbid digit $1$ in every slot. Tool #7 (Identify Subproblems) handles the small endpoint cleanup: the padded count includes $000$ (which is not in our range) but excludes nothing else we want, so a single subtraction fixes it.
Execute — Answer: D
3.OA.A.1 Step 1 - Reframe every number $0, 1, \ldots, 999$ as a $3$-digit string by padding with leading zeros.
- This turns the problem into "how many $3$-character strings over digits $\{0,1,\ldots,9\}$ avoid the digit $1$ in every position?" Each of the $3$ positions independently has $9$ allowed digits (everything except $1$).
💡 "$9$ choices, repeated for each independent slot" is the same equal-groups multiplication idea taught in Grade 3.
5.NBT.B.5 Step 2 Apply the multiplication principle across the $3$ independent positions (hundreds, tens, ones).
💡 Three independent $9$-way choices multiply, just like the Grade 5 standard multi-digit multiplication fluency.
4.OA.A.3 Step 3 - Adjust for the endpoints.
- The $729$ strings include the padded string "$000$", which represents $0$ — but $0$ is not strictly between $1$ and $1000$, so it must be removed.
- The number $1$ is already excluded automatically (the string "$001$" contains a $1$), and $1000$ has four digits so it was never counted.
💡 Counting then peeling off the one bad case at the boundary is the Tool #7 subproblems pattern, exactly the multi-step word-problem reasoning of Grade 4.
4.OA.A.3 Step 4 Match the count to the answer choices.
💡 Choosing the matching option closes the problem.
3.OA.A.1 Reframe every number $0, 1, \ldots, 999$ as a $3$-digit string by padding with l 5.NBT.B.5 Apply the multiplication principle across the $3$ independent positions (hundred 4.OA.A.3 Adjust for the endpoints. The $729$ strings include the padded string "$000$", w 4.OA.A.3 Match the count to the answer choices. Review
Reasonableness: Roughly $\tfrac{9}{10}$ of digits are allowed in each of $3$ slots, so the fraction of $3$-digit strings that avoid $1$ is about $\left(\tfrac{9}{10}\right)^3 = 0.729$. Out of the $1000$ strings from $000$ to $999$, that predicts about $729$ — matching our $728$ exactly after subtracting $000$. The other choices fail this sanity check: $512 = 8^3$ would forbid two digits, $800$ would forbid only fractions of slots, etc.
Alternative: Tool #7 (Identify Subproblems) by digit length. (i) $1$-digit numbers $2$–$9$ without a $1$: there are $8$. (ii) $2$-digit numbers $10$–$99$ without a $1$: tens digit has $8$ choices ($\{2,\ldots,9\}$), ones digit has $9$ choices, giving $8 \times 9 = 72$. (iii) $3$-digit numbers $100$–$999$ without a $1$: hundreds digit has $8$ choices, tens has $9$, ones has $9$, giving $8 \times 9 \times 9 = 648$. Total $= 8 + 72 + 648 = 728$, confirming (D).
CCSS standards used (min grade 5)
3.OA.A.1Interpret products of whole numbers as equal groups (Recognizing that "$9$ allowed digits for each of $3$ slots" is an equal-groups multiplication setup.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Computing the product $9 \times 9 \times 9 = 729$ from the three independent slot choices.)4.OA.A.3Solve multi-step word problems with whole numbers (Subtracting the single boundary case (the padded string $000$) to adjust the count from $729$ to $728$, and selecting the matching choice.)
⭐ Pad every number to three digits, then each slot is a separate $9$-choice pick — that is exactly the equal-groups multiplication you learned in Grade 5, with one small fix at the end.
⭐ Pad every number to three digits, then each slot is a separate $9$-choice pick — that is exactly the equal-groups multiplication you learned in Grade 5, with one small fix at the end.
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