What's a good way to filter data based on values that match the right of the decimal place?

Question

Let's say I have the following data in a decimal (18,9) identity column in a database table:

ID
========
1000.00
1001.23
1002.00
1003.23

I want to write a SQL query that returns only those records that have .230000000 for the right side of the decimal place, such that only the following are returned:

1001.23
1003.23

I know I could do this by converting it to a string a doing a string comparison, but is there a better way that doesn't involve string comparison and is more efficient at handling numeric values?

---

Based on Martin's answer, I think the best solution to use is like so:

Abs(ID - Cast(ID As int))

I added the Abs() function to work with negative numbers.

I ran the following tests to make sure I wasn't getting wierd rounding errors:

Declare @ID decimal(18,9)

Set @ID = 1000.000000000
Select @ID, Abs(@ID - Cast(@ID As int))
-- Returns: 1000.000000000  0.000000000

Set @ID = -1000.000000000
Select @ID, Abs(@ID - Cast(@ID As int))
-- Returns: -1000.000000000 0.000000000

Set @ID = 1000.000000001
Select @ID, Abs(@ID - Cast(@ID As int))
-- Returns: 1000.000000001  0.000000001

Set @ID = -1000.000000001
Select @ID, Abs(@ID - Cast(@ID As int))
-- Returns: -1000.000000001 0.000000001

Set @ID = 1000.999999999
Select @ID, Abs(@ID - Cast(@ID As int))
-- Returns: 1000.999999999  0.999999999

Set @ID = -1000.999999999
Select @ID, Abs(@ID - Cast(@ID As int))
-- Returns: -1000.999999999 0.999999999

Answer 1

SELECT
    [ID]
FROM
    [TABLE]
WHERE
    [ID] - ROUND([ID], 0) = .23

HTH,
Kent

Answer 2

Can you create a persisted computed column

id - cast(id as int)

So you can then index it?

Answer 3

How about this?

SELECT MyCol
FROM MyTable
WHERE MyCol - ROUND(MyCol, 0) = .23