Project Euler Problem 55 Solution


If we take 47, reverse and add, 47+74=12147 + 74 = 121, which is palindromic.

Not all numbers produce palindromes so quickly. For example,

349+943=12921292+2921=42134213+3124=7337\displaystyle \begin{aligned} 349 + 943 & = 1292 \\ 1292 + 2921 & = 4213 \\ 4213 + 3124 & = 7337 \end{aligned}

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.


palindrome :: String -> Bool
palindrome s = s == reverse s

lychrel :: Int -> Bool
lychrel n = not $ any palindrome $ take 50 $ tail (iterate (\x -> show $ (read x) + (read $ reverse x)) (show n))

main :: IO ()
main = print $ length $ filter lychrel [1..10000]
$ ghc -O2 -o lychrel lychrel.hs
$ time ./lychrel
real   0m0.295s
user   0m0.288s
sys    0m0.000s


#!/usr/bin/env python

def is_palindrome(s):
    return s == ''.join(reversed(s))

def is_lychrel(n):
    s = str(n)
    i = 0
    done = False
    while not done:
        if i > 50:
            return True
        s = str(int(s) + int(''.join(reversed(s))))
        i += 1
        if is_palindrome(s):
            done = True
    return False

def main():
    count = 0
    for n in range(10000):
        if is_lychrel(n):
            count += 1

if __name__ == "__main__":
$ time python3
real   0m0.153s
user   0m0.148s
sys    0m0.004s

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