How many 3-digit positive integers are odd and do not contain digit 5?

Question

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?

Answer 1

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 $9$.

Answer 2

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 $\dfrac{4}{9}$.