Q108 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
How many natural numbers are there between 100 and 1000 such that at least one of their digits is 6 ?
- A$258$
- Correct answerB$252$
- C$251$
- D$243$
Solution
The total number of natural numbers between 100 and 1000 is the set of all 3-digit numbers, which ranges from 100 to 999. The total count is $999 - 100 + 1 = 900$.
To find the number of integers that have at least one digit as 6, it is easier to calculate the complement: the number of 3-digit integers that contain no digit equal to 6.
For a 3-digit number $d_1 d_2 d_3$ where $d_1 \in \{1, 2, 3, 4, 5, 7, 8, 9\}$, $d_2 \in \{0, 1, 2, 3, 4, 5, 7, 8, 9\}$, and $d_3 \in \{0, 1, 2, 3, 4, 5, 7, 8, 9\}$, the number of choices for each digit is:
For $d_1$: 8 choices (excluding 0 and 6).
For $d_2$: 9 choices (excluding 6).
For $d_3$: 9 choices (excluding 6).
The total number of 3-digit integers with no digit equal to 6 is $8 \times 9 \times 9 = 648$.
The number of 3-digit integers with at least one digit equal to 6 is the total number of 3-digit integers minus the number of 3-digit integers with no digit equal to 6:
$900 - 648 = 252$.
Answer: $252$
To find the number of integers that have at least one digit as 6, it is easier to calculate the complement: the number of 3-digit integers that contain no digit equal to 6.
For a 3-digit number $d_1 d_2 d_3$ where $d_1 \in \{1, 2, 3, 4, 5, 7, 8, 9\}$, $d_2 \in \{0, 1, 2, 3, 4, 5, 7, 8, 9\}$, and $d_3 \in \{0, 1, 2, 3, 4, 5, 7, 8, 9\}$, the number of choices for each digit is:
For $d_1$: 8 choices (excluding 0 and 6).
For $d_2$: 9 choices (excluding 6).
For $d_3$: 9 choices (excluding 6).
The total number of 3-digit integers with no digit equal to 6 is $8 \times 9 \times 9 = 648$.
The number of 3-digit integers with at least one digit equal to 6 is the total number of 3-digit integers minus the number of 3-digit integers with no digit equal to 6:
$900 - 648 = 252$.
Answer: $252$
Hence, the correct answer is option: 'B'.