The digital sum of a 3-digit number is even if
• all 3 of its digits are even, or
• 2 of its digits are odd and 1 is even
Count the number of 3-digit numbers, all of whose digits are even:
• the first digit cannot be 0, and otherwise has 4 options (2, 4, 6, or 8)
• the second and third digits each have 5 options (2, 4, 6, 8, or 0)
So there are 4 × 5² = 100 such numbers.
Count the number of 3-digit numbers with 2 odd digits and 1 even digit:
• if the first digit is even, then there are 4 options (2, 4, 6, or 8), and the other two digits each have 5 options (1, 3, 5, 7, or 9)
• if the first digit is odd, then there are 5 options (1, 3, 5, 7, or 9); if the second digit is also odd, it has the same 5 options; then the third digit is even and has 5 options (2, 4, 6, 8, or 0)
Swapping the second and third digits in the last case gets the same amount of numbers; we're effectively multiplying by the number of permutations of 2 objects taken 2 at a time, 2!.
So there are 4 × 5² + 2! × 5³ = 350.
Then there is a total of 450 3-digit numbers whose digital sums are even.
