A set of three-digit numbers

I have this problem that's broken into smaller ones.

Concerning a set of 3-digit numbers (between 100 - 999 inclusive):
1) How many are divisible by 5 -- 180
2) How many are not divisible by 5 -- 720
3) How many are divisible by 4 -- 225
4) How many are divisible by 4 or 5
5) How many are divisible by 4 and 5
6) How many are divisible by neither 4 nor 5

I have the first 3. I'm stuck on the rest. I think that for #5, I want to say 405. For #6, I want to say 495. For the third one, I'm really confused.

Thanks for any help

Try some logical reasoning; like:

4,8,12,16,20 : 1/5 of numbers divisible by 4 are also divisible by 5

C'est bon?

tragicallylost said:

Concerning a set of 3-digit numbers (between 100 - 999 inclusive):
4) How many are divisible by 4 or 5
5) How many are divisible by 4 and 5
6) How many are divisible by neither 4 nor 5

Click to expand...

The following function gives the number divisible by n in that range: (BTW it uses the ceiling function.) \(\displaystyle \L D\left( n \right) = \left\lceil {\frac{{999}}{n}} \right\rceil - \left\lceil {\frac{{99}}{n}} \right\rceil\).

4) \(\displaystyle D\left( 4 \right) + D\left( 5 \right) - D\left( {20} \right)\)

5) \(\displaystyle D\left( {20} \right)\)

6) \(\displaystyle 900 - \left[ {D\left( 4 \right) + D\left( 5 \right) - D\left( {20} \right)} \right]\)

Hi. Thanks so much for your help. I know that this is a formula, and I'm still kind of lost. If you don't mind, could you explain some things for me?

Like with question 4, you have to take into account those that are divisible by 4 and by 5. Why those divisible by 20? Aren't those already accounted for?

For question 5, I assumed that you could just add those that are divisible by four and divisible by 5.

One last thing, with question 6, who come you couldn't take the sum of those that are both divisible by four and five, and subtract it from the total number of the set (999-100+1)?

Thanks

tragicallylost said:

you have to take into account those that are divisible by 4 and by 5. Why those divisible by 20? Aren't those already accounted for?

For question 5, I assumed that you could just add those that are divisible by four and divisible by 5.

One last thing, with question 6, who come you couldn't take the sum of those that are both divisible by four and five, and subtract it from the total number of the set (999-100+1)?

Click to expand...

Any number that is divisible by both 4 & 5 must also be divisible by 20. To see that- 30 is divisible by 5 but not by 4 or 20; 36 is divisible by 4 but not 5 or 20; 60 is divisible by 4, 5, & 20. Thus as we count D(4)+D(5) we have counted all the numbers divisible by 20 twice. So we subtract.
You do know about the inclusion/exclusion principle?

Oh, ok. I kind of see it now. No, the inclusion/exclusion principle is the next section ahead. Thanks again for the help!

tragicallylost said:

No, the inclusion/exclusion principle is the next section ahead.0

Click to expand...

Good grief, that textbook has the cart before the horse.

Hopefully you can see why I'm having trouble with this

tragicallylost said:

Hopefully you can see why I'm having trouble with this

Click to expand...

Indeed I can. I should have suspected something was amiss because of the order of the questions. Authors who are really familiar with this material would have asked the last three questions in a different order: 5), 6), then 4). Rewording 6) "not divisible by 4 or 5". Of course, doing so only after introducing inclusion/exclusion. In fact, many textbooks use this sort of question to illustrate that principle.