Add solutions for problems 55-58.

Problem055.hs

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+-- 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.
+--
+-- How many Lychrel numbers are there below ten-thousand?
+
+import Euler
+
+step n = n + fromDigits (reverse $ toDigits n :: [Word8])
+palindrome n = let ds = toDigits n :: [Word8] in ds == reverse ds
+lychrel n = not $ any palindrome $ take 50 $ iterate step (step n)
+main = print $ length $ filter lychrel [1..10000]

Problem056.hs

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+-- Considering natural numbers of the form, ab, where a, b < 100, what is the maximum digital sum?
+import Euler
+main = print $ maximum $ [ sum (toDigits (a^b) :: [Integer]) | a <- [1..100], b <- [1..100] ]

Problem057.hs

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+-- It is possible to show that the square root of two can be expressed as an infinite continued fraction.
+--
+-- √ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
+--
+-- In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?
+import Euler
+import Data.Ratio
+
+nDigits 0 = 0
+nDigits n = 1 + nDigits (n `div` 10)
+
+iterations = iterate (\x -> 1+1/(1+x)) (3%2)
+
+main = print $ length $ filter (\x -> nDigits (numerator x) > nDigits (denominator x)) $ take 1000 iterations

Problem058.hs

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+-- Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.
+--
+-- 37 36 35 34 33 32 31
+-- 38 17 16 15 14 13 30
+-- 39 18  5  4  3 12 29
+-- 40 19  6  1  2 11 28
+-- 41 20  7  8  9 10 27
+-- 42 21 22 23 24 25 26
+-- 43 44 45 46 47 48 49
+--
+-- It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out
+-- of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
+--
+-- If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process
+-- is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below
+-- 10%?
+module Main where
+
+import Euler
+
+isPrime n = n >= 2 && not (any (\p -> n `mod` p == 0) $ takeWhile ((<= n) . (^2)) primes)
+
+-- diagonals of an m by m spiral; m = 2*n + 1.
+allDiags = 1 : zipWith (+) allDiags (concat $ map (replicate 4) [2,4..])
+allDiags' = map isPrime allDiags
+diags' n = take (4*n+1) allDiags'
+
+main = print $ head $ [ 2*n+1 | n <- [1..], (length (filter id (diags' n)) % (4*n+1)) < (1 % 10) ]