How many 3-digit positive integers exist whose digits add up to give an even number

To find the number of 3-digit positive integers whose digits add up to an even number, we can break down the problem into smaller steps:

Step 1: Determine the possible values for the hundreds digit. Since we are looking for 3-digit positive integers, the hundreds digit cannot be zero. Therefore, the possible values for the hundreds digit are 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Step 2: Determine the possible values for the tens and units digits. Since the digits must add up to an even number, we need to consider the possible combinations of even and odd digits for the tens and units places.

For the tens digit, we have two cases: Case 1: The hundreds digit is odd. In this case, the tens digit can be any even number from 0 to 8 (inclusive), since an even number plus an odd number is always odd.

Case 2: The hundreds digit is even. In this case, the tens digit can be any number from 0 to 9 (inclusive), since an even number plus an even number is always even.

For the units digit, we also have two cases: Case 1: The tens digit is odd. In this case, the units digit can be any even number from 0 to 8 (inclusive).

Case 2: The tens digit is even. In this case, the units digit can be any number from 0 to 9 (inclusive).

Step 3: Calculate the total number of possible combinations. To find the total number of 3-digit positive integers, we multiply the number of possibilities for each digit together.

For the hundreds digit, there are 9 possibilities. For the tens digit, there are 5 possibilities (4 even numbers for Case 1 and 5 numbers for Case 2). For the units digit, there are 5 possibilities (4 even numbers for Case 1 and 5 numbers for Case 2).

Therefore, the total number of 3-digit positive integers whose digits add up to an even number is 9 * 5 * 5 = 225.