How many integers from $1$ through $1000$ are not divisible by any one of $4, 5$, and $6$?

Question

I hate to use this resource as an argument settler but my friend and I have come across this question and we cannot agree on an answer.

I got $500$ through the use of inclusion and exclusion and he got $466$ through the use of gathering LCM's and such.

Who is correct?

Answer 1

Inclusion Exclusion makes more sense to me than "the use of gathering LCM's and such"

But you do need LCM to do inclusion exclusion.

There are $1000$ integers. $250$ divisible by $4$ and $200$ by $5$ and $166$ by $6$. There are $1000/20 = 50$ there are divisible by both $4$ and $5$ (divisible by $4$ and $5$ means divisible by $20$). But to be divisible by both $4$ and $6$ means to be divisible by $12$, not $24$ so there are $83$ divisible by both $4$ and $6$. And $33$ by both $6$ and $5$. And to be divisible by all $3$ is to be divisible by $60$ and there are $16$ of those.

So the answer is $1000 - 250 - 200 - 166 + 50 + 83 + 33 - 16 = 534$.

Answer 2

Let A,B,C be the sets of integers between $1$ and $1000$ (inclusive) which are divisible $4,5,6$. We want $|(A\cup B\cup C)^c|$

Now $$|(A\cup B\cup C)|=|A|+|B|+|C|-|A\cap B|-|B\cap C|-|A\cap C|+|A\cap B\cap C|$$

Now find $$|A|=\lfloor\dfrac{1000}{4}\rfloor=?$$ $$|B|=\lfloor\dfrac{1000}{5}\rfloor=?$$ $$|C|=\lfloor\dfrac{1000}{6}\rfloor=?$$ $$|A\cap B|=\lfloor\dfrac{1000}{20}\rfloor=?$$ and so on .....