Understanding the Problem

The question asks us to find the total count of non-negative integers that are strictly less than 1000 and contain the digit '1' at least once. Non-negative integers less than 1000 range from 0 to 999.

Strategy: Complementary Counting

A common strategy for problems asking to count items with a certain property is to use complementary counting. This means we will calculate the total number of items in the set and subtract the number of items that *do not* have the desired property.

In this case:

• Total non-negative integers less than 1000.
• Subtract the count of non-negative integers less than 1000 that *do not* contain the digit '1'.

Calculating Total Numbers

The non-negative integers less than 1000 are 0, 1, 2, ..., 999. There are exactly 1000 such integers.

To make the counting process easier, we can represent all these numbers as 3-digit numbers by padding with leading zeros where necessary (e.g., 0 is 000, 5 is 005, 42 is 042). So, we are considering numbers from 000 to 999.

The total number of possibilities for a 3-digit number, where each digit can be any of the 10 digits (0 through 9), is:

Total numbers = (Number of choices for 1st digit) $\times$ (Number of choices for 2nd digit) $\times$ (Number of choices for 3rd digit)

Total numbers = $10 \times 10 \times 10 = 10^3 = 1000$.

Calculating Numbers Without the Digit '1'

Now, let's count how many of these 3-digit numbers (000 to 999) do *not* contain the digit '1'.

For each of the three digit positions (hundreds, tens, units), we can choose any digit *except* '1'. The available digits are {0, 2, 3, 4, 5, 6, 7, 8, 9}. There are 9 possible choices for each position.

Number of integers without the digit '1' = (Choices for 1st digit) $\times$ (Choices for 2nd digit) $\times$ (Choices for 3rd digit)

Number of integers without the digit '1' = $9 \times 9 \times 9 = 9^3$.

Calculating the value: $9^3 = 729$.

Final Calculation

The number of non-negative integers less than 1000 that *contain* the digit '1' is:

Count = (Total numbers) - (Numbers without the digit '1')

Count = $10^3 - 9^3$

Count = $1000 - 729 = 271$.

Conclusion

The number of non-negative integers less than 1000 that contain the digit 1 is $10^3 - 9^3$.