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

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

The correct is 288.

My idea is, an initial I get the total variety of 3-digit integers that do not save 5, then division it by 2. And because the is a 3-digit integer, the hundreds digit deserve to not be zero.

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



There are four digits the the number can finish with and also be odd, no $\frac92$, i beg your pardon is what her calculation provides -- that is, there are more even numbers there is no a 5 than odd numbers without a five.

More correctly:

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


There is no reason that there are just as many odd integers that perform not save on computer $5$ together there are also integers that do contain 5. The proper fraction is $\dfrac49$.


To answer your inquiry specifically, your idea is not correct because after you eliminate the integers that contain 5, you no longer have a 1:1 proportion of even:odd integers, so you can"t just divide through 2 to get your "number of weird integers that do not save the number 5."


out the the nine digits 0,1,2,3,4,6,7,8, and 9. The digit at hundred ar may be any digit other than 0, any type of of the nine digits have the right to occupy tens place and the unit place can be populated by 1,3,7 and 9. For this reason 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 might 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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