Question

The total number of 3-digit numbers, the sum of whose digits is even, is equal to:

• A. 450
• B. 350
• C. 250
• D. 325

Hint:

When finding the number of favorable outcomes, break down the problem into smaller cases based on the conditions provided.

Answer 1

The Correct Option is A

Step 1: A 3-digit number can be represented as \( abc \), where \( a, b, c \) are the hundreds, tens, and ones digits, respectively.
- \( a \) can be any digit from 1 to 9 (since it’s a 3-digit number), so there are 9 choices for \( a \).
- \( b \) and \( c \) can be any digit from 0 to 9, so there are 10 choices for both \( b \) and \( c \).
Thus, the total number of 3-digit numbers is: \[ 9 \times 10 \times 10 = 900. \] Step 2: For the sum \( a + b + c \) to be even, the sum of the digits must be even. This can happen in two cases:
- Case 1: \( a \) is even, \( b + c \) is even.
- Case 2: \( a \) is odd, \( b + c \) is odd.
In Case 1, \( a \) can be \( 2, 4, 6, 8 \) (4 choices). For \( b + c \) to be even, both \( b \) and \( c \) must be even, so there are 5 choices for \( b \) and 5 choices for \( c \). Therefore, the number of such numbers is: \[ 4 \times 5 \times 5 = 100. \] In Case 2, \( a \) can be \( 1, 3, 5, 7, 9 \) (5 choices). For \( b + c \) to be odd, one of \( b \) or \( c \) must be odd, and the other must be even. There are 5 choices for odd digits and 5 choices for even digits. The number of such numbers is: \[ 5 \times 5 \times 5 = 125. \] Thus, the total number of 3-digit numbers where the sum of the digits is even is: \[ 100 + 125 = 450. \]