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Euler problems/111 to 120

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== [http://projecteuler.net/index.php?section=problems&id=111 Problem 111] ==
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Do them on your own!
Search for 10-digit primes containing the maximum number of repeated digits.
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Solution:
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<haskell>
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import Control.Monad (replicateM)
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-- All ways of interspersing n copies of x into a list
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intr :: Int -> a -> [a] -> [[a]]
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intr 0 _ y      = [y]
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intr n x (y:ys) = concat
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                  [map ((replicate i x ++) . (y :)) $ intr (n-i) x ys
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                      | i <- [0..n]]
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intr n x _      = [replicate n x]
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-- All 10-digit primes containing the maximal number of the digit d
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maxDigits :: Char -> [Integer]
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maxDigits d = head $ dropWhile null
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              [filter isPrime $ map read $ filter ((/='0') . head) $
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              concatMap (intr (10-n) d) $
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              replicateM n $ delete d "0123456789"
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                  | n <- [1..9]]
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problem_111 = sum $ concatMap maxDigits "0123456789"
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=112 Problem 112] ==
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Investigating the density of "bouncy" numbers.
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Solution:
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<haskell>
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isIncreasing' n p
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    | n == 0 = True
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    | p >= p1 = isIncreasing' (n `div` 10) p1
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    | otherwise = False
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    where
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    p1 = n `mod` 10
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isIncreasing :: Int -> Bool
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isIncreasing n = isIncreasing' (n `div` 10) (n `mod` 10)
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isDecreasing' n p
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    | n == 0 = True
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    | p <= p1 = isDecreasing' (n `div` 10) p1
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    | otherwise = False
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    where
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    p1 = n `mod` 10
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isDecreasing :: Int -> Bool
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isDecreasing n = isDecreasing' (n `div` 10) (n `mod` 10)
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isBouncy n = not (isIncreasing n) && not (isDecreasing n)
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nnn=1500000
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num150 =length [x|x<-[1..nnn],isBouncy x]
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p112 n nb
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    | fromIntegral nnb / fromIntegral n >= 0.99 = n
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    | otherwise = prob112' (n+1) nnb
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    where
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    nnb = if isBouncy n then nb + 1 else nb
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problem_112=p112 (nnn+1) num150
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=113 Problem 113] ==
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How many numbers below a googol (10100) are not "bouncy"?
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Solution:
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<haskell>
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import Array
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mkArray b f = listArray b $ map f (range b)
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digits = 100
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inc = mkArray ((1, 0), (digits, 9)) ninc
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dec = mkArray ((1, 0), (digits, 9)) ndec
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ninc (1, _) = 1
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ninc (l, d) = sum [inc ! (l-1, i) | i <- [d..9]]
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ndec (1, _) = 1
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ndec (l, d) = sum [dec ! (l-1, i) | i <- [0..d]]
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problem_113 = sum [inc ! i | i <- range ((digits, 0), (digits, 9))]
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              + sum [dec ! i | i <- range ((1, 1), (digits, 9))]
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              - digits*9 -- numbers like 11111 are counted in both inc and dec
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              - 1 -- 0 is included in the increasing numbers
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</haskell>
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Note: inc and dec contain the same data, but it seems clearer to duplicate them.
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it is another way to solution this problem:
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<haskell>
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binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
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prodxy x y=product[x..y]
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problem_113=sum[binomial (8+a) a+binomial (9+a) a-10|a<-[1..100]]
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=114 Problem 114] ==
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Investigating the number of ways to fill a row with separated blocks that are at least three units long.
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Solution:
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<haskell>
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-- fun in p115
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problem_114=fun 3 50
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=115 Problem 115] ==
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Finding a generalisation for the number of ways to fill a row with separated blocks.
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Solution:
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<haskell>
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binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
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prodxy x y=product[x..y]
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fun m n=sum[binomial (k+a) (k-a)|a<-[0..div (n+1) (m+1)],let k=1-a*m+n]
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problem_115 = (+1)$length$takeWhile (<10^6) [fun 50 i|i<-[1..]]
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=116 Problem 116] ==
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Investigating the number of ways of replacing square tiles with one of three coloured tiles.
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Solution:
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<haskell>
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binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
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prodxy x y=product[x..y]
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f116 n x=sum[binomial (a+b) a|a<-[1..div n x],let b=n-a*x]
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p116 x=sum[f116 x a|a<-[2..4]]
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problem_116 = p116 50
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=117 Problem 117] ==
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Investigating the number of ways of tiling a row using different-sized tiles.
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Solution:
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<haskell>
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fibs5 = 0 : 0 :1: 1:zipWith4 (\a b c d->a+b+c+d) fibs5 a1 a2 a3
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    where
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    a1=tail fibs5
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    a2=tail a1
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    a3=tail a2
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p117 x=fibs5!!(x+2)
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problem_117 = p117 50
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=118 Problem 118] ==
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Exploring the number of ways in which sets containing prime elements can be made.
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Solution:
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<haskell>
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digits = ['1'..'9']
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-- possible partitions voor prime number sets
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-- leave out patitions with more than 4 1's
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-- because only {2,3,5,7,..} is possible
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-- and the [9]-partition because every permutation of all
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-- nine digits is divisable by 3
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test xs
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    |len>4=False
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    |xs==[9]=False
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    |otherwise=True
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    where
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    len=length $filter (==1) xs
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parts = filter test $partitions  9
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permutationsOf [] = [[]]
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permutationsOf xs = [x:xs' | x <- xs, xs' <- permutationsOf (delete x xs)]
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combinationsOf  0 _ = [[]]
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combinationsOf  _ [] = []
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combinationsOf  k (x:xs) =
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    map (x:) (combinationsOf (k-1) xs) ++ combinationsOf k xs
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priemPerms [] = 0
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priemPerms ds =
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    fromIntegral . length . filter (isPrime . read) . permutationsOf $ ds
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setsums [] 0 = [[]]
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setsums [] _ = []
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setsums (x:xs) n
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    | x > n    = setsums xs n
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    | otherwise = map (x:) (setsums (x:xs) (n-x)) ++ setsums xs n
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partitions n = setsums (reverse [1..n]) n
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fc :: [Integer] -> [Char] -> Integer
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fc (p:[]) ds = priemPerms ds
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fc (p:ps) ds =
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    foldl fcmul 0 . combinationsOf p $ ds
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    where
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    fcmul x y
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        | np y == 0 = x
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        | otherwise = x + np y * fc ps (ds \\ y)
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        where
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        np = priemPerms
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-- here is the 'imperfection' correction method:
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-- make use of duplicate reducing factors for partitions
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-- with repeating factors, f.i. [1,1,1,1,2,3]:
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-- in this case 4 1's -> factor = 4!
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-- or for [1,1,1,3,3] : factor = 3! * 2!
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dupF :: [Integer] -> Integer
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dupF = foldl (\ x y -> x * product [1..y]) 1 . map (fromIntegral . length) . group
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main = do
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    print . sum . map (\x -> fc x digits `div` dupF x) $ parts
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problem_118 = main
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=119 Problem 119] ==
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Investigating the numbers which are equal to sum of their digits raised to some power.
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Solution:
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<haskell>
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import Data.List
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digits n
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{-  123->[3,2,1]
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-}
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    |n<10=[n]
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    |otherwise= y:digits x
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    where
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    (x,y)=divMod n 10
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problem_119 =sort [(a^b)|
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    a<-[2..200],
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    b<-[2..9],
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    let m=a^b,
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    let n=sum$digits m,
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    n==a]!!29
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</haskell>
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== [http://projecteuler.net/index.php?section=problems&id=120 Problem 120] ==
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Finding the maximum remainder when (a − 1)n + (a + 1)n is divided by a2.
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Solution:
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<haskell>
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fun m=div (m*(8*m^2-3*m-5)) 3
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problem_120 = fun 500
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</haskell>
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Revision as of 21:45, 29 January 2008

Do them on your own!