In the following sum the digits 0 to 9 have all been used, O = Odd, E = Even, zero is even and the top row's digits add to 9. Can you determine each digit?

Hint: The first digit in each row is even, so the first column can only be 2 + 8, 4 + 6, 4 + 8, 6 + 8.

Answer: 423 + 675 = 1098.

Remembering that:

even + even = even
odd + odd = even
even + odd = odd

To discuss individual letters it's easiest to represent the sum as:

A B C
D E F +
--------
G H I J

A and D are both even, and can only be 2 + 8, 4 + 6, 4 + 8, 6 + 8, so the carry can only be 1, so G = 1.

Since column 2 is even + odd = odd there can be no carry from column 1 (since even + odd + 1 is always even). Therefore neither C nor F (both odd) can be 7 or 9 (otherwise there would be a carry), so C and F are 3 and 5 (but we don't yet know which is which), therefore J = 8.

Because A + D is even there can't be a carry from column 2, therefore E can't be 9 as this would force a carry. Therefore I = 9. Hence B can't be 0. Therefore H = 0.

The last remaining odd number makes E = 7. Making B = 2.

Therefore A and D are 4 and 6 (but we don't yet know which is which).

Since the top row's digits have to add to 9 the top number must be 423.