it is a GRE question. And it has actually been reply here. Yet I still desire to ask that again, simply to understand why ns am wrong.

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

The correct is 288.

My idea is, first I gain the total variety of 3-digit integers that do not contain 5, then division it through 2. And also because it is a 3-digit integer, the hundreds digit can not be zero.

So, I have actually (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, not \$frac92\$, i beg your pardon is what your calculation provides -- that is, there are an ext even numbers without a 5 than weird numbers there is no a five.

More correctly:

\$8 * 9 * 4 = 72 * 4 = 288\$, as the an initial digit can be any type of of \$1,2,3,4,6,7,8,9\$, the second any yet \$5\$, and also the third must it is in \$1,3,7,\$ or \$9\$.

There is no factor that there are simply as many odd integers that execute not save \$5\$ together there are also integers that do contain 5. The proper portion is \$dfrac49\$.

To prize your question specifically, her idea is no correct because after you eliminate the integers the contain 5, you no longer have a 1:1 proportion of even:odd integers, so friend can"t merely divide by 2 to get your "number of weird integers that perform not save on computer the digit 5."

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

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

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