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?
 A: There are four digits that the number can end with and be odd, not $\frac{9}{2}$, which is what your calculation uses -- that is, there are more even numbers without a five than odd numbers without a five.
More correctly:
$8 * 9 * 4 = 72 * 4 = 288$, as the first digit can be any of $1,2,3,4,6,7,8,9$, the second any but $5$, and the third must be $1,3,7,$ or $9$.
A: There is no reason that there are just as many odd integers that do not contain $5$ as there are even integers that do contain 5. The proper fraction is $\dfrac{4}{9}$.
A: To answer your question specifically, your idea is not correct because after you eliminate the integers that contain 5, you no longer have a 1:1 ratio of even:odd integers, so you can't simply divide by 2 to get your "number of odd integers that do not contain the digit 5."
A: out of the nine digits 0,1,2,3,4,6,7,8, and 9. The digit at hundred place may be any digit other than 0, any of the nine digits can occupy tens place and the unit place can be occupied by 1,3,7 and 9. Thus the required number of three digits odd numbers will be 8*9*4=288
A: (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) $\rightarrow
(8) (9) (4) = 288$
