Question

I am doing question 12 of project euler where I must find the first triangle number with 501 divisors. So I whipped up this with Haskell:

divS n = [ x | x <- [1..(n)], n `rem` x == 0 ]
tri n = (n* (n+1)) `div` 2
divL n = length (divS (tri n))
answer = [ x | x <- [100..] ,  501 == (divL x)]

The first function finds the divisors of a number.

The second function calculates the nth triangle number

The 3rd function finds the length of the list that are the divisors of the triangle number

The 4th function should return the value of the triangle number which has 501 divisors.

But so far this run for a while without returning a result. Is the answer very large or do I need some serious optimisation to make this work in a realistic amount of time?

Answer 1

You need to use properties of divisor function: http://en.wikipedia.org/wiki/Divisor_function

Notice that n and n + 1 are always coprime, so that you can get d(n * (n + 1) / 2) by multiplying previously computed values.