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

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

The correct is 288.

My idea is, first I gain the total number of 3-digit integers that execute not save on computer 5, then divide it through 2. And because the is a 3-digit integer, the hundreds digit deserve to not it is in zero.

So, I have actually (8*9*9)/2 = 324. Why this idea is not the correct?


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There are four digits the the number can finish with and be odd, no $\frac92$, which is what your calculation supplies -- that is, over there are much more even numbers there is no a 5 than odd numbers there is no a five.

More correctly:

$8 * 9 * 4 = 72 * 4 = 288$, as the very first digit deserve to 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 $9$.


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There is no factor that over there are simply as many odd integers that perform not contain $5$ together there are even integers that carry out contain 5. The proper fraction is $\dfrac49$.


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To answer your concern specifically, your idea is no correct because after you remove the integers that contain 5, you no longer have actually a 1:1 proportion of even:odd integers, so friend can"t simply divide by 2 to get your "number of odd integers that perform not contain the digit 5."


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out of the nine digits 0,1,2,3,4,6,7,8, and also 9. The number at hundred place may be any kind of digit various other than 0, any type of of the nine digits deserve to occupy 10s place and also the unit place have the right to be populated by 1,3,7 and also 9. Hence the required number of three digits odd numbers will certainly be 8*9*4=288


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


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