Find the sum of all the multiples of 3 or 5 below 1000

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.

The previously posted answer isn't correct. The statement of the problem is to sum the multiples of 3 and 5 below 1000, not up to and equal 1000. The correct answer is \begin{eqnarray} \sum_{k_{1} = 1}^{333} 3k_{1} + \sum_{k_{2} = 1}^{199} 5 k_{2} - \sum_{k_{3} =1}^{66} 15 k_{3} = 166833 + 99500 - 33165 = 233168, \end{eqnarray} where we have the used the identity \begin{eqnarray} \sum_{k = 1}^{n} k = \tfrac{1}{2} n(n+1). \end{eqnarray}

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.