How many 3-digit positive integers are odd and do not contain digit 5? It is a GRE question. And it has been answered here. But I still want to ask it again, just to know why I am wrong.
The correct is 288.
My idea is, first I get the total number of 3-digit integers that do not contain 5, then divide it by 2. And because it is a 3-digit integer, the hundreds digit can not be zero.
So, I have (8*9*9)/2 = 324. Why this idea is not the correct?

(Hundreds) (Tens) (Units), 
Units could be $(1, 3, 7, 9) \rightarrow 4$ numbers, 
Tens could be $(0, 1, 2, 3, 4, 6, 7, 8, 9)\rightarrow 9$ numbers,
Hundreds could be $(1, 2, 3, 4, 6, 7, 8, 9) \rightarrow 8$ numbers,
(Hundreds) (Tens) (Units) $\rightarrow
(8) (9) (4) = 288$
