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.
 A: Well the main equation is already given above. The only question which give me trouble is that why I have to subtract the sum of 15?! Well, the answer is, 15 can be evenly divide by both 3 & 5. So the products of 15 can also be divided by those number as well! So, when you adding the numbers with Sum Of Three & Sum Of Five there are some numbers(i.e. 15,30,45,60....) which are available at both SUMMATION. So, you have to subtract at least once from the total sum to get the answer!
Hope this helps someone like me:) !!
A: 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}
A: 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.
A: 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.
