that is a GRE question. And also it has been reply here. Yet I still desire to ask the again, just to know why i am wrong.

You are watching: How many 3-digit positive integers are odd

The exactly is 288.

My idea is, very first I obtain the total number of 3-digit integers that carry out not save on computer 5, then division it by 2. And because the 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?



There are 4 digits the the number can finish with and be odd, no $\frac92$, which is what her calculation uses -- the 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 very first digit can be any kind of of $1,2,3,4,6,7,8,9$, the 2nd any yet $5$, and also the 3rd must be $1,3,7,$ or $9$.


There is no reason that there are simply as numerous odd integers that do not save $5$ as there are also integers that do contain 5. The proper fraction is $\dfrac49$.


To answer your concern specifically, her idea is no correct due to the fact that after you get rid of the integers the contain 5, friend no longer have actually a 1:1 ratio of even:odd integers, so friend can"t simply divide through 2 to gain your "number of odd integers that carry out not save the digit 5."


out the the ripe digits 0,1,2,3,4,6,7,8, and also 9. The number at hundred ar may be any type of digit various other than 0, any of the ripe digits can occupy tens place and the unit place can be populated by 1,3,7 and also 9. Therefore the required variety of three number odd numbers will certainly be 8*9*4=288

(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) $\rightarrow(8) (9) (4) = 288$

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