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I need to find the number of natural numbers between 1 and 1000 that are divisible by 3, 5 or 7.

I know that given a set of numbers, 1 ... n, the number of numbers divisible by d is given by $\lfloor \frac{n}{d} \rfloor$

So if we let $A$ be the set of numbers divisible by 3, $B$ be the set of numbers divisible by 5 and $C$ be the set of numbers divisible by 7, then the size of the union of those sets is going to be the number required.

However, to work out $A \cup B \cup C$, I need to know $A \cap B$ (and $A \cap C$ etc).

My only thought was that to work out $A \cap B$, I could use $\lfloor \frac{1000}{3 \times 5} \rfloor$, but I am unsure if this misses out some numbers. Is that the correct way to do it or not? If no, can anyone suggest the correct way.

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up vote 6 down vote accepted

In general, you have to use $\left\lfloor\frac{1000}{[a,b]}\right\rfloor$, where $[a,b]$ is the least common multiple of $a$ and $b$.

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That's right. A number is divisible by 3 and 5 if and only if it is divisible by 15.

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You can be sure that it does not miss any number. It comes out as an multiplication because you are looking for the common multiple of 2 prime numbers.

It would be the same for the other intersections that you need. $$(A\cap B), (A \cap B \cap C) $$

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divisibility by 3 or 5 or by both means: N(3 or 5)=N(3) + N(5) -N(3 and 5) N(3 and 5) bcoz remove all common from nos. divisible by 3 and 5

N(3)=1000/3=333 N(5)=1000/5=200 N(3 and 5)=66 so 333+200-66=467

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Welcome to MSE. You may wish to have a look at our tutorial to learn how to typeset math here. Also, why did you choose to answer such an old question? Your answer is hard to read and doesn't add anything to the others already present. (Not to mention that, being one answer already accepted, it is unlikely that other people will see your answer, due to how the site works.) – A.P. Apr 16 '15 at 18:16

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