1. ## Counting

Suppose we have 6000 oranges.

If we throw out:

Every 3rd orange because it's too small,
Every 4th orange because it's too orange and,
Every 10th orange because it's too bruised,

How many determine how many oranges are perfect (meaning how many would we have left after doing the above)?

2. Hello,
Originally Posted by DiscreteW
Suppose we have 6000 oranges.

If we throw out:

Every 3rd orange because it's too small,
Every 4th orange because it's too orange and,
Every 10th orange because it's too bruised,

How many determine how many oranges are perfect (meaning how many would we have left after doing the above)?
Do it the rough way

There are 600 multiples of 10 between 1 and 6000.
There are 1500 multiples of 4 between 1 and 6000.
There are 2000 multiples of 3 between 1 and 6000.

To the multiples of 4, remove the ones that are multiples of 10. For four multiples of 4, there is one that is a multiple of 10 ---> keep 3/4 of them, that is to say 1125.

To the multiples of 3, remove the ones that are multiples of 10 and multiples of 4.
- multiples of 10 ---> keep 9/10 (for 10 multiples of 3, there is one that is a multiple of 10) of them, that is to say 1800.
- multiples of 4, that are not multiples of 10..................... This part is the most difficult in my opinion, so try it out (it's not that difficult)

If you have problems with countings, do it with small numbers, so that you can see the reasoning

3. Originally Posted by Moo
Hello,

Do it the rough way

There are 600 multiples of 10 between 1 and 6000.
There are 1500 multiples of 4 between 1 and 6000.
There are 2000 multiples of 3 between 1 and 6000.

To the multiples of 4, remove the ones that are multiples of 10. For four multiples of 4, there is one that is a multiple of 10 ---> keep 3/4 of them, that is to say 1125.

To the multiples of 3, remove the ones that are multiples of 10 and multiples of 4.
- multiples of 10 ---> keep 9/10 (for 10 multiples of 3, there is one that is a multiple of 10) of them, that is to say 1800.
- multiples of 4, that are not multiples of 10..................... This part is the most difficult in my opinion, so try it out (it's not that difficult)

If you have problems with countings, do it with small numbers, so that you can see the reasoning
Hmm you lost me on the multiples of 4, remove the ones that are multiples of 10 part...

We know that 4 goes into 20 -- 5 times- so, wouldn't we want to remove every 1 in 5 oranges.

Then, for multiples of 3, remove multiples of 10... so wouldn't we want to remove every 1 in 30?

And then for multiples of 4 that ARENT multiples of 10... wouldn't we want to remove every 4 out of 5?

Is there an easier way to do this? Does my thinking make sense?

Obviously I see what it comes down to is removing all the multiples of 3, 4, and 10 (and there will be some overlapping I realize)

4. Originally Posted by DiscreteW
Hmm you lost me on the multiples of 4, remove the ones that are multiples of 10 part...

We know that 4 goes into 20 -- 5 times- so, wouldn't we want to remove every 1 in 5 oranges.

Then, for multiples of 3, remove multiples of 10... so wouldn't we want to remove every 1 in 30?

And then for multiples of 4 that ARENT multiples of 10... wouldn't we want to remove every 4 out of 5?

Is there an easier way to do this? Does my thinking make sense?

Obviously I see what it comes down to is removing all the multiples of 3, 4, and 10 (and there will be some overlapping I realize)
Tried going through your reasoning and still get the same result (my reasoning). Anyone want to work through it w/ me or show me an easier way?

5. $\begin{gathered} D_{3} \mbox{ is the multiples of 3} \hfill \\
D_{4} \mbox{ is the multiples of 4} \hfill \\ D_{10} \mbox{ is the multiples of 10} \hfill \\ \end{gathered}$
.
The number of multiples of 3, 4, or 10 is given by the inclusion/exclusion principle:
$\left| {D_3 \cup D_4 \cup D_{10} } \right| = \left| {D_3 } \right| + \left| {D_4 } \right| + \left| {D_{10} } \right| - \left| {D_{12} } \right| - \left| {D_{30} } \right| - \left| {D_{40} } \right| + \left| {D_{120} } \right|$

Now calculate the various numbers.
The answer you want is the difference in the total and that number.