Counting divisibility from 1 to 1000 Of the integers $1, 2, 3, ..., 1000$, how many are not divisible by $3$, $5$, or $7$?
The way I went about this was
$$\text{floor}(1000/3) + \text{floor}(1000/5) + \text{floor}(1000/7)-\text{floor}(1000/(3\cdot5)) - \text{floor}(1000/(3\cdot7))-\text{floor}(1000/(5\cdot7))+\text{floor}(1000/(3\cdot5\cdot7))$$
which resulted in $543$ and then I subtracted that from $1000$ to get $457$.
I do not have an answer key so I was wondering if that was the right approach to the question.
Any help or insight would be appreciated!

The idea is to use the Inclusion/Exclusion principle.  Let us first count how many numbers are divisible by $3$, $5$, or $7$.  Let set $X$ be the set of all such numbers.
Let $A$ = {Numbers divisible by $3$}
Let $B$ = {Numbers divisible by $5$}
Let $C$ = {Numbers divisible by $7$}
By the inclusion/exclusion principle:
$|X| = |A| + |B| + |C| - |A\cap B| - |A\cap C| - |B\cap C| + |A \cap B \cap C|$.  Then, of course, the answer you're looking for will be $1000-|X|$.
I'll leave the implementation of this up to you.  :)
http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
