Project Euler – Problem 55 Solution

Yan Cui

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Problem

If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.

Not all numbers produce palindromes so quickly. For example,

349 + 943 = 1292,

1292 + 2921 = 4213

4213 + 3124 = 7337

That is, 349 took three iterations to arrive at a palindrome.

Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).

Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is 4994.

How many Lychrel numbers are there below ten-thousand?

NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.

Solution

	open System.Linq
	// checks if the number n is palindromic
	let isPalindromic n =
	let charArray = n.ToString().ToCharArray()
	let revCharArray = Array.rev charArray
	charArray.SequenceEqual(revCharArray)
	// reverse a number, e.g. 1234 -> 4321
	let reverse n =
	bigint.Parse(n.ToString().ToCharArray() |> Array.rev |> Array.map string |> Array.reduce (+))
	// checks if a number is lychrel, i.e. not produce palindromic sum in 50 iterations
	let rec isLychrelRec i (number:bigint) =
	let sum = number + (reverse number)
	if isPalindromic sum then false
	else if i >= 49 then true else isLychrelRec (i+1) sum
	// curry the recursive isLychrelRec function and initialize with 1
	let isLychrel = isLychrelRec 1
	let answer = [1I..10000I] |> List.filter isLychrel |> List.length

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