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.
The multiples of 3 are 3,6,9,12,15,18,21,24,27,30,....
The multiples of 5 are 5,10,15,20,25,30,35,40,45,....
The intersection of these two sequences is 15,30,45,...
The sum of the first numbers 1+2+3+4+...+n is n(n+1)/2.
The sum of the first few multiples of k, say k+2k+3k+4k+...+nk must be kn(n+1)/2.
Now you can just put these ingredients together to solve the problem.
Since we are asked to look for numbers below 1000, we shall look at numbers up to the number 999.
To find n, use 999/3 = 333 + remainder, 999/5 = 199 + remainder, 999/15 = 66 + remainder, by using a*(m*(m+1)/2) , where m=n/a. here a is 3 or 5 or 15, and n is 999 or 1000 but 999 is best, and then sum multiples of 3: $3((333)*(333+1)/2) = 166833$ plus multiples of 5: $5((199)*(199+1)/2) = 99500$; and subtract multiples of 15 $15((66)+(66+1)/2 )= 33165$ to get 233168.