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# How many positive integers are there that are divisible by 2, 5, or 7, and<=4900 Watch

1. I defined 3 sets

Used the inclusion-exclusion principle;

and got 3220 which I'm not sure is right. My question is, is there a quicker, nicer way of doing this? The question also states 4900 = (2.5.7)^2 which I haven't used.
2. 4900/2 gives the amount of numbers that are divisible by 2 <=4900
4900/5 gives the amount of numbers that are divisible by 5 <= 4900
Then take away the value of 4900/10 since adding 4900/2 + 4900/5 gives duplicate numbers that are both divisible by 2 and 5.

Then you can use this principle to find numbers that are divisible by 7 too.

I got the value of 3150?
3. (Original post by 0x2a)
4900/2 gives the amount of numbers that are divisible by 2 <=4900
4900/5 gives the amount of numbers that are divisible by 5 <= 4900
Then take away the value of 4900/10 since adding 4900/2 + 4900/5 gives duplicate numbers that are both divisible by 2 and 5.

Then you can use this principle to find numbers that are divisible by 7 too.

I got the value of 3150?
That is the inclusion-exclusion principle. I think you need to add 70 onto your answer because in the process of taking away the duplicates, you remove the 4900/(2.5.7)=70 numbers that are divisible by all three.
4. (Original post by TheJ0ker)
That is the inclusion-exclusion principle. I think you need to add 70 onto your answer because in the process of taking away the duplicates, you remove the 4900/(2.5.7)=70 numbers that are divisible by all three.
Yea, figured that out after
5. Programming says 3220 is the answer. I suppose that you missed the multiples of (2*5*7) that were taken off twice.

And the inclusion-exclusion principle basically gives the method you suggested, right 0x2a? So I wouldn't say that it's quicker or nicer...

Yeah, it would be a bit off if this was the most elegant way of solving this problem... :/

EDIT: I'm slow.
6. (Original post by aznkid66)
Programming says 3220 is the answer. I suppose that you missed the multiples of (2*5*7) that were taken off twice.

And the inclusion-exclusion principle basically gives the method you suggested, right 0x2a? So I wouldn't say that it's quicker or nicer...

Yeah, it would be a bit off if this was the most elegant way of solving this problem... :/

EDIT: I'm slow.
Yeah I've thought about it a lot today and I can't think of another method although I'm almost certain there is a nice way of doing it.
7. Another way of doing it is to look at all the numbers divisible by 2,5 and 7 all the way up to and then multiply this answer by 70

So there are 35 numbers that are divisible by 2, there are 14 that are divisible by 5, however 7 of these are even and already counted. So there are 35+7 = 42 numbers out of the first 70 which are divisible by either 2 or 5. So now we need to add on the multiples of 7 which are not even and not divisible by 5. These are 21, 49 and 63, so there are 45 that are divisible by 2,5 or 7. This doesn't include the terms divisible by all three which adds another term. So in total there are 46*70 = 3220.

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Updated: April 11, 2013
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