(a) 450
Out of the 10 digits 0, 1, 2, 3, ..., 9, five digits, i.e., 0, 2, 4, 6, 8 are even and five digits, i.e., 1, 3, 5, 7 and 9 are odd.
The sum of the three digits D1, D2, D3 of the number D1D2D3 will be even if :
(i) All the three digits are even:
∴ Number of such numbers = 4 × 5 × 5 = 100
(Here the position D1 cannot be occupied by 0)
(ii) One of the digits is even and the rest two are odd :
(a) D1 is even, D2 is odd, D3 is odd
∴ Number of such numbers = 4 × 5 × 5 = 100
(Again since D1 ≠ 0, so only 4 choices for D1)
(b) D1 is odd, D2 is even, D3 is odd
∴ Number of such numbers = 5 × 5 × 5 = 125
(c) D1 is odd, D2 is odd, D3 is even
∴ Number of such numbers = 5 × 5 × 5 = 125
∴ Total number of required numbers
= 100 + 100 + 125 + 125 = 450.
