How do you find “How many 3 digit positive integers are odd and do not contain the digit 5?”?

The answer is 288, but I don’t know how it was done. Please show work.

Top 2 Answers

A few days ago

TitoBob

Favorite Answer

In each group of 10 numbers, beginning at 101, there are five odd ones. Exclude the one ending in 5, and you have four of them. That will work for all the groups of ten except the fifties, so that makes nine groups of four or 36 numbers within each hundred. You will also have to leave out all the 500’s, so you end up with eight sets of 36 numbers up to 999 (you can’t count the first 100, because they only have one or two digits), which comes to 288 numbers in all.

A few days ago

aguynamedbrett

That last guy’s not helping you out. Here – I was looking for the same question (I’m assuming we’re both looking at the same GRE study guide) and found this guy’s answer –

Let’s assume the first digit is neither 0 nor 5. There are 8 numbers which could be used for the hundreds place.

For the number in the tens place there are a total of 9 numbers from which to choose (0, 1, 2, 3, 4, 6, 7, 8, 9)

Finally, the number in the ones place must be odd and it cannot be a five. There are just 4 such digits. (1, 3, 7, 9)

Thus the number of possible numbers is:

8 Ã 9 Ã 4 = 288

there ya go man, good luck

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How do you find “How many 3 digit positive integers are odd and do not contain the digit 5?”?

How do you find “How many 3 digit positive integers are odd and do not contain the digit 5?”?

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