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

You are watching: How many 3 digit positive integers are odd and do not contain the digit 5

The exactly is 288.

My idea is, an initial I get the total variety of 3-digit integers that do not contain 5, then division it by 2. And 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 no the correct?

There are four digits that the number can finish with and be odd, not $\\frac92$, i beg your pardon is what your calculation uses -- the is, there are much more even numbers there is no a 5 than odd numbers there is no a five.

More correctly:

$8 * 9 * 4 = 72 * 4 = 288$, together the very first digit can be any of $1,2,3,4,6,7,8,9$, the 2nd any yet $5$, and the third must be $1,3,7,$ or $9$.

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

To answer your inquiry specifically, your idea is no correct since after you eliminate the integers that contain 5, friend no longer have a 1:1 proportion of even:odd integers, so friend can\"t merely divide by 2 to gain your \"number of weird integers that execute not save on computer the number 5.\"

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

(Hundreds) (Tens) (Units), Units can be $(1, 3, 7, 9) \\rightarrow 4$ numbers, Tens can 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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How countless $10$-digit numbers (allowing initial digit to be zero) in which only $5$ that the $10$ possible digits are represented?