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

Question

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.

Answer 1

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}

Answer 2

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.

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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.