TITLE: Find the sum of all the multiples of 3 or 5 below 1000
QUESTION [47 upvotes]: How to solve this problem, I can not figure it out:
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.

REPLY [31 votes]: First of all, stop thinking on the number $1000$ and turn your
 attention to the number $990$ instead. If you solve the problem
 for $990$ you just have to add $993, 995, 996$ & $999$ to it for
 the final answer. This sum is $(a)=3983$
Count all the #s divisible by $3$: From $3$... to $990$ there are
 $330$ terms. The sum is $330(990+3)/2$, so $(b)=163845$ 
Count all the #s divisible by $5$: From $5$... to $990$ there are
 $198$ terms. The sum is $198(990+5)/2$, so $(c)=98505$
Now, the GCD (greatest common divisor) of $3$ & $5$ is $1$, so the
 LCM (least common multiple) should be $3\times 5 = 15$.
This means every number that divides by $15$ was counted twice,
 and it should be done only once. Because of this, you have an
 extra set of numbers started with $15$ all the way to $990$ that
 has to be removed from (b)&(c).
Then, from $15$... to $990$ there are $66$ terms and their sum is
 $66(990+15)/2$, so $(d)=33165$
The answer for the problem is: $(a)+(b)+(c)-(d) = 233168$
Simple but very fun problem.