Solving the Project Euler Problems – Problem 1

After stumbling across an article on Medium by Bennett Garner, I discovered the Project Euler problems and that it would be a good idea to showcase my skills in solving them in a variety of ways, as a good starting point to get a code portfolio going, so here goes with – as natural a starting point as any – problem 1!

All of the Project Euler Problems can be found at

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.

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There were 2 ways I considered to tackle this:

  • Loop from 1 to 1000 (or as in the code, n) and check each number for dividing by 3 and 5 respectively, adding them to the result if either is true
  • Reduce the number of loops by 47% and run our loop n/3 + n/5 times (for those interested, 1/3 + 1/5 = 8/15, so this reduction percentage works for any integer), by dividing n by 3, running that loop, adding 3 times each number to the result – which takes division out of the loop as well – then doing the same again, only now having to check (on a smaller loop) if our result was already added as being a multiple of 3.

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