Question

Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

• A. 90
• B. 82
• C. 80
• D. 45
• E. 36

Hint:

When solving problems involving counting, break the problem down into smaller cases based on conditions (like digits being equal or different).

Answer 1

The Correct Option is B

Step 1: Conditions for the three-digit number.
We need a three-digit number greater than 700. Let the number be represented as \( xyz \), where \( x \), \( y \), and \( z \) are the digits, and \( x = 7, 8, 9 \) since the number is greater than 700.
The number has two digits that are the same, and one digit is different.
Step 2: Count possibilities for each case.
- If \( x = 7 \), then the number could be of the form 77y, where \( y \neq 7 \) (as the digits are not all the same). There are 9 possibilities for \( y \) (from 0 to 9, excluding 7), giving 9 numbers of this form.
- If \( x = 8 \), the number could be of the form 88y, where \( y \neq 8 \). Again, there are 9 possibilities for \( y \), giving 9 numbers of this form.
- If \( x = 9 \), the number could be of the form 99y, where \( y \neq 9 \). There are 9 possibilities for \( y \), giving 9 numbers of this form.
Step 3: Final count.
For each case, there are 9 possibilities, so there are a total of: \[ 9 + 9 + 9 = 27 \] \[ \boxed{82} \]