Project Euler 5: Multiplicity

The fifth problem from Project Euler asks for “the smallest number that is evenly divisible by all of the numbers from 1 to 20”. The key, for me, was to paraphrase that as “the least common multiple of 1 to 20”.

The least common multiple (lcm) of two natural numbers is their product divided by their greatest common divisor (gcd, a well-known lab rat). So, the stock definitions of lcm and gcd, along with one quick fold, are all we need.

  def divisibleByAll(lo: Int, hi: Int) = {
    def gcd(a: Int, b: Int) = gcd1(a max b, a min b)
    def gcd1(a: Int, b: Int): Int = if (b == 0) a else gcd(b, a % b)
    def lcm(a: Int, b: Int) = a * (b / gcd(a, b))
    (lo to hi).foldLeft(1)(lcm(_, _))

There is one subtlety in this code; gcd is only there to insure that the (natural) arguments to gcd1 are correctly ordered (larger first). That is only needed once; if a >= b, then a % b cannot be larger than b, so the recursion in gcd1 is guaranteed to maintain the larger-first argument ordering.

Finally, this design encounters integer overflow at 1..24, changing the inner functions to work with Long gets us up to 1..42 before long overflow kicks in.

It should be obvious that the answers for 1..21 and 1..22 are the same as for 1..20, but I had to do a double-take while scanning for the limits. Duh.

Trackbacks are closed, but you can post a comment.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: