Problem 22 · 2009 AMC 8
Medium
Counting & Probability
casework-on-digit-count
How many whole numbers between 1 and 1000 do not contain the digit 1?
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Answer: D — 728.
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Hint 1 of 2
Counting by independent digit-choices is the move: each digit slot can be filled freely, and 'no digit 1' just means each slot avoids one value. Multiply the choices per slot.
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Hint 2 of 2
Be careful at the boundaries: the FIRST digit of a number can't be 0 (or it's shorter), and remember 1000 itself contains a 1. Splitting by number of digits keeps the leading-zero rule straight.
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Approach: multiply allowed choices per digit, split by length
- Forbidden digit is 1. A non-leading slot may be any of {0,2,3,4,5,6,7,8,9} = 9 choices; a leading slot also bans 0, leaving {2,…,9} = 8 choices.
- 1-digit: 8. 2-digit: 8 × 9 = 72. 3-digit: 8 × 9 × 9 = 648. (1000 has a 1, so it's out.)
- Total: 8 + 72 + 648 = 728.
- Why this transfers: 'how many numbers avoid digit d' is a per-slot multiplication — just mind the leading-zero rule and the top endpoint.
Another way — pad to 3 digits and subtract:
- Write every number 1–999 as a 3-digit string 001…999, allowing leading zeros. With no '1' anywhere, each of the 3 slots has 9 allowed digits: 9³ = 729 strings (this counts 000).
- Drop 000 (it's not in our range) and note 1000 has a 1: 729 − 1 = 728. Padding turns three cases into one clean power.
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