Question

How many natural numbers are there between 100 and 1000 such that at least one of their digits is 6?

• A. 243
• B. 251
• C. 252
• D. 258

Hint:

For “at least one” type digit problems, use complement method: total cases minus cases with no required digit.

Answer 1

The Correct Option is C

Step 1: Identify the range.
Numbers between 100 and 1000 are three-digit numbers.
So total numbers are:
\[ 999-100+1=900 \]

Step 2: Use complement method.
Instead of counting numbers having digit 6 directly, count numbers having no digit 6.
Then subtract from total three-digit numbers.

Step 3: Count hundreds place choices without digit 6.
Hundreds place can be:
\[ 1,2,3,4,5,7,8,9 \]
So number of choices:
\[ 8 \]

Step 4: Count tens place choices without digit 6.
Tens place can be any digit from 0 to 9 except 6.
So number of choices:
\[ 9 \]

Step 5: Count units place choices without digit 6.
Units place can also be any digit from 0 to 9 except 6.
So number of choices:
\[ 9 \]

Step 6: Find numbers having no digit 6.
\[ 8 \times 9 \times 9 = 648 \]

Step 7: Find numbers having at least one digit 6.
\[ 900 - 648 = 252 \]
\[ \boxed{252} \]