that is a GRE question. And also it has been answer here. But I still want to ask the again, just to recognize 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, very first I obtain the total number of 3-digit integers that carry out not save 5, then divide it by 2. And also because that is a 3-digit integer, the hundreds digit have the right 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 that the number can finish with and be odd, no \$frac92\$, which is what your calculation offers -- the is, there are much more even numbers without 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 however \$5\$, and the third must it is in \$1,3,7,\$ or \$9\$. There is no reason that there are simply as plenty of odd integers that carry out not contain \$5\$ together there are even integers that perform contain 5. The proper portion is \$dfrac49\$. To prize your question specifically, your 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 proportion of even:odd integers, so girlfriend can"t simply divide by 2 to gain your "number of odd integers that perform not contain the number 5." out the the ripe digits 0,1,2,3,4,6,7,8, and also 9. The number at hundred location may be any kind of digit other than 0, any of the ripe digits can occupy 10s place and the unit place can be occupied by 1,3,7 and 9. Hence the required variety of three digits odd numbers will certainly be 8*9*4=288

(Hundreds) (Tens) (Units), Units can be \$(1, 3, 7, 9) ightarrow 4\$ numbers, Tens can 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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