To solve the common problem associated with the sequence 1-100 (counting numbers containing the digit "1"), here's the breakdown:

Step 1: Count numbers with digit "1"

• 1-9: Only the number "1" → 1 number.
• 10-19: All 10 numbers (10,11,...,19) have the digit "1" → 10 numbers.
• 20-99: Numbers where the units digit is "1" (21,31,...,91) → 8 numbers (tens place: 2-9, 8 values).
• 100: Contains the digit "1" in the hundreds place → 1 number.

Total

1 + 10 + 8 + 1 = 20

Answer: 20 (assuming the question is "how many numbers from 1 to 100 have at least one digit '1'").

If the question was "how many times does the digit '1' appear?", the answer would be 21 (10 times in units place, 10 in tens place, 1 in hundreds place). But the most common question here is the count of numbers, so 20 is the intended answer.

Final Answer
$\boxed{20}$

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