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.
We are asked to add two series of numbers. The first is 3,6,9,...,996,999. This can be represented as 3 *(1+2+3+...+332+333), or more generically, 3*(1+2+3+...+(n-1)+n), where n=(Limit-1)\3) with \ representing an integer division operator. We use -1 since the requirement is "below 1000" and not "less than or equal to 1000."
The sum of all the numbers from 1 to n is given by n*(n+1)/2. So, the required sum collapses to 3*n*(n+1)/2 where n=(1000-1)\3=333.
One can similarly compute the 2nd series of numbers 5*(1+2+...+(n-1)+n) where n=(Limit - 1) \ 5.
There is one more issue to address. The 2 series have some number of overlapping elements. These are numbers that are multiples of both 3 and 5, i.e., 15, 30, 45, etc. These numbers have effectively been double-counted. So, one set of these numbers should be subtracted. Of course, these numbers themselves constitute the series 15*(1+2+...+n) where n= (Limit-1) \ 15.
Put the three series together to solve Problem 1 without any computer code!
