it is a GRE question. And also it has been answered here. But I still desire to ask it again, simply to understand why i am wrong.

You are watching: How many three-digit positive integers are there

The correct is 288.

My idea is, very first I acquire the total number of 3-digit integers that perform not contain 5, then division it through 2. And also because that is a 3-digit integer, the hundreds digit deserve to not it is in zero.

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



There are four digits the the number can finish with and be odd, no $frac92$, i beg your pardon is what her calculation uses -- that is, there are much more even numbers there is no a five than odd numbers without a five.

More correctly:

$8 * 9 * 4 = 72 * 4 = 288$, as the first digit can be any kind of of $1,2,3,4,6,7,8,9$, the 2nd any however $5$, and the 3rd must be $1,3,7,$ or $9$.


There is no factor that there are simply as countless odd integers that carry out not save $5$ as there are even integers that carry out contain 5. The proper fraction is $dfrac49$.


To answer your concern specifically, her idea is no correct since after you get rid of the integers the contain 5, you no longer have a 1:1 proportion of even:odd integers, so girlfriend can"t simply divide by 2 to acquire your "number of odd integers that perform not save the number 5."


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

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

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