samskivert: Euler 042

01 June 2008

Problem 042:

object Euler42 extends EulerApp {
  def wvalue (word :Seq[Char]) = word.foldLeft(0)((s, l) => (s + (l - 'A' + 1)))
  def tri (n :Int) = (n*(n+1))/2
  println(readwords("words.txt").map(wvalue).filter(v => (v == tri(Math.sqrt(2*v)))).length);
}

This one is kind of cheating. I discovered (using the terminology tri = (n * (n+1))/2) that floor(sqrt(2*tri)) == n for at least the first 10,000 values of n. This isn't hugely surprising because 2*tri = (n*n + n), so the square root of that is always going to be a bit bigger than n, and floor of that could well take us right to the integer we need. Who knew?

It may be true for larger values of n but BigInt doesn't have sqrt() so I didn't feel like going through the pain of checking. Certainly the biggest n we encounter in words.txt is way smaller than 10,000 (it is in fact 19).

©1999–2015 Michael Bayne