How many 3-digit positive integers are odd and do not contain the digit “5”?
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The lowest 3-digit number is 100, since if the first digit were 0, it would only be 2-digits. Now think of the number as three boxes to fill: □ □ □
The leftmost box can be 1 to 9, but not 5. 8 choices.
The middle box can be 0 to 9, but not 5, 9 choices.
The rightmost box can only be 1, 3, 7, or 9 if the number is odd and not 5. 4 choices.
So the count is 8•9•4 = 72•4 =288 such 3-digit positive numbers.
The leftmost box can be 1 to 9, but not 5. 8 choices.
The middle box can be 0 to 9, but not 5, 9 choices.
The rightmost box can only be 1, 3, 7, or 9 if the number is odd and not 5. 4 choices.
So the count is 8•9•4 = 72•4 =288 such 3-digit positive numbers.
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Using the convention that numbers of more than one digit do not begin with the digit "0" the first digit can be any of 8 (excluding 0 and 5), the second digit can be any of 9 (excluding 5), and the final digit can be any of 4 (excluding 5 and the even digits).
This means 8×9×4 = 288.
This means 8×9×4 = 288.
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405