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

problem on project euler site

my solution on github

Let’s start with the intuitive solution to this problem. Loop over the range 1 to 999, and use a boolean statement to decide if you are going to add that integer to the sum.

def reduce_three_and_five(n)
  (1...n).reduce(0) { |m, n| n % 3 == 0 || n % 5 == 0 ? m + n : m }

The above method takes the range from 1 to n and adds the integer to the sum if it is divisible by 3 or 5. Can you see why this might be wasteful? For just those numbers, it runs almost instantaneously, but what if we increase the range to 10,000,000?

       user     system      total        real
   1.160000   0.010000   1.170000 (  1.163298)

Our runtime is increasing proportionally to the number of integers that we loop over. This is referred to as linear time or O(n). If we find a way to calculate the answer without a loop, the time that it takes will not depend on the size of the loop. This is constant time or O(1).

Let’s suppose that we only had to add the multiples of three up to 100. We know that

The sum of a series of multiples of three is equal to 3 times the sum of an equally long integer sequence. It just so happens that we have a formula for the sum of sequential integers.

Where is the number of integers to sum. We can use this to make a method that will return this sum for us.

def integer_sum(n)
  n * (n + 1) / 2

We know from before that if we want the multiples of 3 up to 100, we would multiply 3 times the series of integers up to 33. This can be generalized to

Where is the number that the series is divisible by, and is the maximum number in that series integer divided by , and is the formula from before. In ruby, this becomes

def multiples_sum(n, l)
  m = l/n
  n * integer_sum(m)

We now have a handy method that will give us the sum of multiples of n up to a maximum l, and in O(1) time. All that is left is to call this three times in our solution method:

def math_three_and_five(n)
  multiples_sum(3, n-1) +
  multiples_sum(5, n-1) -
  multiples_sum(15, n-1)

We add the multiples of three under a number to the multiples of five under a number. Since the multiples of three overlap the multiples of five, we subtract the multiples of fifteen that were duplicated. A quick test shows that this gets the same answer as our reduce method, but without iteration.