Euler problems/61 to 70: Difference between revisions

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== [http://projecteuler.net/index.php?section=view&id=61 Problem 61] ==
== [http://projecteuler.net/index.php?section=problems&id=61 Problem 61] ==
Find the sum of the only set of six 4-digit figurate numbers with a cyclic property.
Find the sum of the only set of six 4-digit figurate numbers with a cyclic property.


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<haskell>
<haskell>
import Data.List
import Data.List
 
triangle  = [n*(n+1)`div`2  | n <- [1..]]
permute [] = [[]]
square    = [n^2            | n <- [1..]]
permute xs = concatMap (\x -> map (x:) $ permute $ delete x xs) xs
pentagonal = [n*(3*n-1)`div`2 | n <- [1..]]
   
hexagonal = [n*(2*n-1)       | n <- [1..]]
figurates n xs = extract $ concatMap (gather (map poly xs)) $ map (:[]) $ poly n
heptagonal = [n*(5*n-3)`div`2 | n <- [1..]]
  where gather [xs] (v:vs)  
octagonal  = [n*(3*n-2)       | n <- [1..]]
          = let v' = match xs v
 
            in if v' == [] then [] else map (:v:vs) v'
triangle4  = fourDigs triangle
        gather (xs:xss) (v:vs)  
square4    = fourDigs square
          = let v' = match xs v
pentagonal4 = fourDigs pentagonal
            in if v' == [] then [] else concatMap (gather xss) $ map (:v:vs) v'
hexagonal4  = fourDigs hexagonal
        match xs (_,v) = let p = (v `mod` 100)*100 in sublist (p+10,p+100) xs
heptagonal4 = fourDigs heptagonal
        sublist (s,e) = takeWhile (\(_,x) -> x<e) . dropWhile (\(_,x) -> x<s)
octagonal4  = fourDigs octagonal
         link ((_,x):xs) = x `mod` 100 == (snd $ last xs) `div` 100
 
        diff (x:y:xs) = if fst x /= fst y then diff (y:xs) else False
fourDigs = takeWhile (<10000) . dropWhile (<1000)
        diff [x]      = True
 
        extract = filter diff . filter link
solve = do
        poly m = [(n, x) | (n, x) <- zip [1..] $ takeWhile (<10000)  
    (l1:l2:l3:l4:l5:l6:_) <- permute [triangle4, square4, pentagonal4, hexagonal4, heptagonal4, octagonal4]
                                              $ scanl (+) 1 [m-1,2*m-3..],
    a <- l1
                                    1010 < x, x `mod` 100 > 9]
    let m = filter (g a) l2
    b <- m
problem_61 = sum $ map snd $ head $ concatMap (figurates 3) $ permute [4..8]
    let n = filter (g b) l3
    c <- n
    let o = filter (g c) l4
    d <- o
    let p = filter (g d) l5
    e <- p
    let q = filter (g e) l6
    f <- q
    if g f a then return (sum [a,b,c,d,e,f]) else fail "burp"
    where
         g x y = x `mod` 100 == y `div` 100
 
permute        :: [a] -> [[a]]
permute []      = [[]]
permute list = concat $ map (\(x:xs) -> map (x:) (permute xs)) (take (length list) (unfoldr (\x -> Just (x, tail x ++ [head x])) list))
 
problem_61 = head $ solve
</haskell>
</haskell>


== [http://projecteuler.net/index.php?section=view&id=62 Problem 62] ==
== [http://projecteuler.net/index.php?section=problems&id=62 Problem 62] ==
Find the smallest cube for which exactly five permutations of its digits are cube.
Find the smallest cube for which exactly five permutations of its digits are cube.


Line 52: Line 35:
<haskell>
<haskell>
import Data.List
import Data.List
 
import Data.Maybe
cubes = [(x, show $ x^3) | x <- [1..100000]]
a = map (^3) [0..10000]
 
b = map (sort . show) a
problem_62 = f3 $ head $ head $ sortBy shf $ filter l5 $ groupBy g $ sortBy ss $ map sd cubes
c = filter ((==5) . length) . group . sort $ b
    where
Just d = elemIndex (head (head c)) b
        sd (a, b) = (a, sort b)
problem_62 = toInteger d^3
        shf a b = compare (fst $ head a) (fst $ head b)
        ss a b = compare (snd a) (snd b)
        g a b = (snd a) == (snd b)
        l5 a = length a == 5
        f3 a = (fst a)^3
</haskell>
</haskell>


== [http://projecteuler.net/index.php?section=view&id=63 Problem 63] ==
== [http://projecteuler.net/index.php?section=problems&id=63 Problem 63] ==
How many n-digit positive integers exist which are also an nth power?
How many n-digit positive integers exist which are also an nth power?


Solution:
Solution:
Since d<sup>n</sup> has at least n+1 digits for any d≥10, we need only consider 1 through 9.  If d<sup>n</sup> has fewer than n digits, every higher power of d will also be too small since d < 10.  We will also never have n+1 digits for our nth powers.  All we have to do is check d<sup>n</sup> for each d in {1,...,9}, trying n=1,2,... and stopping when d<sup>n</sup> has fewer than n digits.
<haskell>
<haskell>
problem_63 = length . concatMap (takeWhile (\(n,p) -> n == nDigits p))
problem_63=length[x^y|x<-[1..9],y<-[1..22],y==(length$show$x^y)]
            $ [powers d | d <- [1..9]]
    where powers d = [(n, d^n) | n <- [1..]]
          nDigits n = length (show n)
</haskell>
</haskell>


== [http://projecteuler.net/index.php?section=view&id=64 Problem 64] ==
== [http://projecteuler.net/index.php?section=problems&id=64 Problem 64] ==
How many continued fractions for N ≤ 10000 have an odd period?
How many continued fractions for N ≤ 10000 have an odd period?


Solution:
Solution:
<haskell>
<haskell>
import Data.List (findIndex)
import Data.List
 
fraction n = (a0, 1, (-a0)) : nextDigits n 1 (-a0)
problem_64  =length $ filter solve $ [2..9999] \\ (map (^2) [2..100])
    where
        a0 = firstDigit n
solve n = even $ length $ cont n 0 1
 
nextDigits n num den
cont :: Int -> Int -> Int -> [Int]
    | n - (den^2) == 0 = []
cont r n d = m : rest
    | otherwise = (an, den', num') : nextDigits n den' num'
    where
        a0 = firstDigit n
        den' = (n - (den^2)) `div` num
        an = (a0 + abs den) `div` den'
        num' = abs den - (an * den')
 
firstDigit n = floor $ sqrt $ fromInteger n
 
period [] = 0
period xs = period' [] xs
     where
     where
        period' _ [] = 0
    m = (truncate (sqrt (fromIntegral r)) + n) `div` d
        period' acc (x:xs) = case findIndex (x==) acc of
    a = n - d * m
                                Just i -> toInteger(length acc - i)
    rest | d == 1 && n /= 0 = []
                                Nothing -> period' (acc ++ [x]) xs
        | otherwise = cont r (-a) ((r - a ^ 2) `div` d)
 
problem_64 = length $ filter odd $ map (period . fraction) [1..10000]
</haskell>
</haskell>


== [http://projecteuler.net/index.php?section=view&id=65 Problem 65] ==
== [http://projecteuler.net/index.php?section=problems&id=65 Problem 65] ==
Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.
Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.


Solution:
Solution:
<haskell>
<haskell>
import Data.Char
import Data.Ratio
import Data.Ratio
e = 2 : concat [ [1, 2*i, 1] | i <- [1..] ]
fraction [x] = x%1
fraction (x:xs) = x%1 + 1/(fraction xs)


problem_65 = dsum . numerator . contFrac . take 100 $ e
problem_65 = sum $ map digitToInt $ show $ numerator $ fraction $ take 100 e
    where dsum 0 = 0
          dsum n = let ( d, m ) = n `divMod` 10 in m + ( dsum d )
          contFrac = foldr1 (\x y -> x + 1/y)
          e = 2 : 1 : insOnes [2,4..]
          insOnes (x:xs) = x : 1 : 1 : insOnes xs
</haskell>
</haskell>


== [http://projecteuler.net/index.php?section=view&id=66 Problem 66] ==
== [http://projecteuler.net/index.php?section=problems&id=66 Problem 66] ==
Investigate the Diophantine equation x<sup>2</sup> − Dy<sup>2</sup> = 1.
Investigate the Diophantine equation x<sup>2</sup> − Dy<sup>2</sup> = 1.


Solution:
Solution:
Problem solved using the continuous fractions method (reused from problem 64).
<haskell>
<haskell>
import Data.List (inits, maximumBy)
intSqrt :: Integral a => a -> a
import Data.Ratio ((%), numerator, denominator)
intSqrt n
 
     | n < 0 = error "intSqrt: negative n"
fraction :: Integer -> [Integer]
     | otherwise = f n
fraction n = a0 : nextDigits n 1 (-a0)
    where
        a0 = firstDigit n
 
firstDigit :: Integer -> Integer
firstDigit n = floor $ sqrt $ fromInteger n
 
nextDigits :: Integer -> Integer -> Integer -> [Integer]
nextDigits n num den
     | n - (den^2) == 0 = []
     | otherwise = an : nextDigits n den' num'
     where
     where
         a0 = firstDigit n
         f x | y < x = f y
        den' = (n - (den^2)) `div` num
            | otherwise = x
        an = (a0 + abs den) `div` den'
            where y = (x + (n `quot` x)) `quot` 2
        num' = abs den - (an * den')
problem_66 =  
 
    snd$maximum [ (x,d) |  
nonSquares :: [Integer]
    d <- [1..1000],
nonSquares = [x|x <- [2..1000], let s = floor $ sqrt $ fromInteger x, s*s /= x]
    let b = intSqrt d,
 
    b*b /= d, -- d can't be a perfect square
minDiophantineX :: Integer -> Integer
     let (x,_) = pell d b b
minDiophantineX d = numerator $ head $ filter isDiophantine $ map (calc) $ drop 1 $ inits $ fraction d
    ]
     where
        calc x = foldr (\a b -> fromInteger a + 1/b) ((fromInteger $ last x)%1) $ init x
        isDiophantine (r) = (numerator r)^2 - d * (denominator r)^2 == 1
 
maxDiophantine :: [(Integer, Integer)]
maxDiophantine = [(d,minDiophantineX d)|d <- nonSquares]


problem_66 :: Integer
pell d wd b = piter d wd b 0 1 0 1 1 0
problem_66 = fst $ maximumBy (\(_,a) (_,b) -> compare a b) maxDiophantine
piter d wd b i c l k m n
    | cn == 1 = (x, y)
    | otherwise = piter d wd bn (i+1) cn k u n v
    where
    yb = (wd+b) `div` c
    bn = yb*c-b
    cn = (d-(bn*bn)) `div` c
    yn  | i == 0 = wd
        | otherwise = yb
    u = k*yn+l -- u/v is the i-th convergent of sqrt(d)
    v = n*yn+m
    (x,y)  | odd (i+1) = (u*u+d*v*v, 2*u*v)
            | otherwise = (u,v)  
</haskell>
</haskell>


== [http://projecteuler.net/index.php?section=view&id=67 Problem 67] ==
== [http://projecteuler.net/index.php?section=problems&id=67 Problem 67] ==
Using an efficient algorithm find the maximal sum in the triangle?
Using an efficient algorithm find the maximal sum in the triangle?


Solution:
Solution:
<haskell>
<haskell>
problem_67 = do
problem_67 = readFile "triangle.txt" >>= print . solve . parse
    src <- readFile "triangle.txt"
 
    print $ head $ foldr1 g $ parse src
parse :: String -> [[Int]] -- restrict output type for 'read' function
    where
parse = map (map read . words) . lines
        parse :: String -> [[Int]]
        parse s = map ((map read).words) $ lines s
        f x y z = x + max y z
        g xs ys = zipWith3 f xs ys $ tail ys


solve = head . foldr1 step -- reverse pairewise addition from bottom to top
step [] _ = [] -- returen empty list to avoid eception warning
step (x:xs) (y:z:zs) = x + max y z : step xs (z:zs)
</haskell>
</haskell>


== [http://projecteuler.net/index.php?section=view&id=68 Problem 68] ==
Alternatively, one could add lists pairewise from top to bottom. However, this would require a check on maximum in the list, after the final step of additions.
 
== [http://projecteuler.net/index.php?section=problems&id=68 Problem 68] ==
What is the maximum 16-digit string for a "magic" 5-gon ring?
What is the maximum 16-digit string for a "magic" 5-gon ring?


Solution:
Solution:
<haskell>
<haskell>
import Data.List (delete, group, sort)
import Data.List
 
permute []     = [[]]
numberSets :: Int -> [(Int, Int, Int)]
permute list =  
numberSets n = [(a,b,c)| a <- [1..10], b <- delete a [1..10], let c = n - a - b, c > 0, c < 11, a /= c, b /= c]
     concatMap (\(x:xs) -> map (x:) (permute xs))
 
    (take (length list)  
nextSets :: (Int, Int, Int) -> [(Int, Int, Int)]
    (unfoldr (\l@(x:xs) -> Just (l, xs ++ [x])) list))
nextSets (a,b,c) = filter f $ numberSets s
problem_68 =
     where
    maximum $ map (concatMap show) poel
        s = a + b + c
        f x = follows (a, b, c) x
 
follows :: (Int, Int, Int) -> (Int, Int, Int) -> Bool
follows (_,_,c) (_,b',_) = c == b'
 
lowest :: (Ord a) => [a] -> [a]
lowest xs = minimum $ take 6 $ iterate rotate xs
    where
        rotate [] = []
        rotate (y:ys) = ys ++ [y]
 
solutions :: [[(Int, Int, Int)]]
solutions = do
    n <- [13..27]
    a <- numberSets n
    b <- nextSets a
    c <- nextSets b
    d <- nextSets c
    e <- nextSets d
    if follows e a then return [a,b,c,d,e] else fail "not cyclic"
 
magic :: (Ord a) => [a] -> [(a, a, a)] -> Bool
magic acc [] = (length $ group $ sort acc) == 10
magic acc ((a,b,c):xs) = magic (acc++[a,b,c]) xs
 
problem_69 :: [Char]
problem_69 = maximum $ filter (\x -> length x == 16) $ map (toNum . lowest) $ filter (magic []) solutions
     where
     where
        toNum [] = []
    gon68 = [1..10]
         toNum ((a,b,c):xs) = show a ++ show b ++ show c ++ toNum xs
    knip = (length gon68) `div` 2
    (is,e:es) = splitAt knip gon68
    extnodes = map (e:) $ permute es
    intnodes = map (\(p:ps) -> zipWith (\ x y -> [x, y])
         (p:ps) (ps++[p])) $ permute is
    poel = [ concat hs |
            uitsteeksels <- extnodes,
            organen <- intnodes,
            let hs = zipWith (:) uitsteeksels organen,
            let subsom = map sum hs,
            length (nub subsom) == 1 ]
</haskell>
</haskell>


== [http://projecteuler.net/index.php?section=view&id=69 Problem 69] ==
== [http://projecteuler.net/index.php?section=problems&id=69 Problem 69] ==
Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.
Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.


Solution:
Solution:
<haskell>
<haskell>
import Data.Ratio
{-phi(n) = n*(1-1/p1)*(1-1/p2)*...*(1-1/pn)
import Data.List
n/phi(n) = 1/(1-1/p1)*(1-1/p2)*...*(1-1/pn)
 
(1-1/p) will be minimal for a small p and 1/(1-1/p) will then be maximal
primePowerFactors n = rle (takeFactors n primes)
-}
    where rle = map (\xs -> (head xs, length xs)) . group
primes=[2,3,5,7,11,13,17,19,23]
          takeFactors n (p:ps)
problem_69=
            | n == 1         = []
    maximum [c|
            | p * p > n      = [n]
    b<-tail $ inits primes,
            | n `mod` p == 0 = p : takeFactors (n `div` p) (p:ps)
    let c=product b,
            | otherwise      = takeFactors n ps
    c<10^6
 
    ]
eulerTotient n = product (map (\(p,i) -> p^(i-1) * (p-1)) factors)
    where factors = primePowerFactors n
 
problem_69 = snd . maximum . map (\n -> (n % eulerTotient n, n)) $ [1..1000000]
</haskell>
</haskell>


Note: credit for arithmetic functions is due to [http://www.polyomino.f2s.com/ David Amos].
Note: credit for arithmetic functions is due to [http://www.polyomino.f2s.com/ David Amos].


== [http://projecteuler.net/index.php?section=view&id=70 Problem 70] ==
== [http://projecteuler.net/index.php?section=problems&id=70 Problem 70] ==
Investigate values of n for which φ(n) is a permutation of n.
Investigate values of n for which φ(n) is a permutation of n.


Solution:
Solution:
<haskell>
<haskell>
import Data.List (minimumBy, group, sort)
import Data.List
 
import Data.Function
primes :: [Int]
isPerm a b = null $ show a \\ show b
primes = 2 : filter (l1 . primeFactors) [3,5..]
flsqr n x=x<(floor.sqrt.fromInteger) n
     where
pairs n1 =
        l1 (_:[]) = True
     fst . minimumBy (compare `on` fn) $ [(m,pm)|a<-gena,b<-genb,let m=a*b,n>m,let pm=m-a-b+1,isPerm m pm]
        l1      _ = False
 
primeFactors :: Int -> [Int]
primeFactors n = factor n primes
     where
     where
        factor _ [] = []
    n=fromInteger n1
        factor m (p:ps) | p*p > m        = [m]
    gena = dropWhile (flsqr n)$ takeWhile (flsqr (2*n))  primes
                        | m `mod` p == 0 = p : factor (m `div` p) (p:ps)
     genb = dropWhile (flsqr (n `div` 2))$ takeWhile (flsqr n) primes
                        | otherwise      = factor m ps
    fn (x,px) = fromIntegral x / (fromIntegral px)  
 
primePowerFactors :: Int -> [(Int, Int)]
primePowerFactors n = map (\x -> (head x, length x)) $ (group . primeFactors) n
 
phi :: Int -> Int
phi n = product (map f $ primePowerFactors n)
     where
        f (p, l) = (p-1)*(p^(l-1))
 
isPermutation :: (Show a) => a -> a -> Bool
isPermutation a b = (sort $ show a) == (sort $ show b)
 
phiPerms :: [(Int, Int)]
phiPerms = [(n, p)| n <- [2..10000000], let p = phi n, isPermutation n p]


problem_70 :: Int
problem_70= pairs (10^7)
problem_70 = fst $ minimumBy (\(a,b) (a',b') -> compare (fromIntegral a/fromIntegral b) (fromIntegral a'/fromIntegral b')) phiPerms
</haskell>
</haskell>

Latest revision as of 10:20, 18 May 2022

Problem 61

Find the sum of the only set of six 4-digit figurate numbers with a cyclic property.

Solution:

import Data.List
 
permute [] = [[]]
permute xs = concatMap (\x -> map (x:) $ permute $ delete x xs) xs
 
figurates n xs = extract $ concatMap (gather (map poly xs)) $ map (:[]) $ poly n
  where gather [xs] (v:vs) 
          = let v' = match xs v
            in if v' == [] then [] else map (:v:vs) v'
        gather (xs:xss) (v:vs) 
          = let v' = match xs v
            in if v' == [] then [] else concatMap (gather xss) $ map (:v:vs) v'
        match xs (_,v) = let p = (v `mod` 100)*100 in sublist (p+10,p+100) xs
        sublist (s,e) = takeWhile (\(_,x) -> x<e) . dropWhile (\(_,x) -> x<s)
        link ((_,x):xs) = x `mod` 100 == (snd $ last xs) `div` 100
        diff (x:y:xs) = if fst x /= fst y then diff (y:xs) else False
        diff [x]      = True
        extract = filter diff . filter link
        poly m = [(n, x) | (n, x) <- zip [1..] $ takeWhile (<10000) 
                                               $ scanl (+) 1 [m-1,2*m-3..], 
                                     1010 < x, x `mod` 100 > 9]
 
problem_61 = sum $ map snd $ head $ concatMap (figurates 3) $ permute [4..8]

Problem 62

Find the smallest cube for which exactly five permutations of its digits are cube.

Solution:

import Data.List
import Data.Maybe
a = map (^3) [0..10000]
b = map (sort . show) a
c = filter ((==5) . length) . group . sort $ b
Just d = elemIndex (head (head c)) b
problem_62 = toInteger d^3

Problem 63

How many n-digit positive integers exist which are also an nth power?

Solution:

problem_63=length[x^y|x<-[1..9],y<-[1..22],y==(length$show$x^y)]

Problem 64

How many continued fractions for N ≤ 10000 have an odd period?

Solution:

import Data.List
 
problem_64  =length $ filter solve $ [2..9999] \\ (map (^2) [2..100])
 
solve n = even $ length $ cont n 0 1
 
cont :: Int -> Int -> Int -> [Int]
cont r n d = m : rest
    where
    m = (truncate (sqrt (fromIntegral r)) + n) `div` d
    a = n - d * m
    rest | d == 1 && n /= 0 = []
         | otherwise = cont r (-a) ((r - a ^ 2) `div` d)

Problem 65

Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.

Solution:

import Data.Char
import Data.Ratio
 
e = 2 : concat [ [1, 2*i, 1] | i <- [1..] ]
 
fraction [x] = x%1
fraction (x:xs) = x%1 + 1/(fraction xs)

problem_65 = sum $ map digitToInt $ show $ numerator $ fraction $ take 100 e

Problem 66

Investigate the Diophantine equation x2 − Dy2 = 1.

Solution:

intSqrt :: Integral a => a -> a
intSqrt n
    | n < 0 = error "intSqrt: negative n"
    | otherwise = f n
    where
        f x | y < x = f y 
            | otherwise = x
            where y = (x + (n `quot` x)) `quot` 2
problem_66 = 
    snd$maximum [ (x,d) | 
    d <- [1..1000],
    let b = intSqrt d,
    b*b /= d, -- d can't be a perfect square
    let (x,_) = pell d b b 
    ]

pell d wd b = piter d wd b 0 1 0 1 1 0
piter d wd b i c l k m n 
    | cn == 1 = (x, y)
    | otherwise = piter d wd bn (i+1) cn k u n v
    where 
    yb = (wd+b) `div` c
    bn = yb*c-b
    cn = (d-(bn*bn)) `div` c
    yn  | i == 0 = wd
        | otherwise = yb
    u = k*yn+l -- u/v is the i-th convergent of sqrt(d)
    v = n*yn+m
    (x,y)   | odd (i+1) = (u*u+d*v*v, 2*u*v)
            | otherwise = (u,v)

Problem 67

Using an efficient algorithm find the maximal sum in the triangle?

Solution:

problem_67 = readFile "triangle.txt" >>= print . solve . parse

parse :: String -> [[Int]] -- restrict output type for 'read' function
parse = map (map read . words) . lines

solve = head . foldr1 step -- reverse pairewise addition from bottom to top
step [] _ = [] -- returen empty list to avoid eception warning
step (x:xs) (y:z:zs) = x + max y z : step xs (z:zs)

Alternatively, one could add lists pairewise from top to bottom. However, this would require a check on maximum in the list, after the final step of additions.

Problem 68

What is the maximum 16-digit string for a "magic" 5-gon ring?

Solution:

import Data.List
permute []      = [[]]
permute list = 
    concatMap (\(x:xs) -> map (x:) (permute xs))
    (take (length list) 
    (unfoldr (\l@(x:xs) -> Just (l, xs ++ [x])) list))
problem_68 = 
    maximum $ map (concatMap show) poel 
    where
    gon68 = [1..10]
    knip = (length gon68) `div` 2
    (is,e:es) = splitAt knip gon68
    extnodes = map (e:) $ permute es
    intnodes = map (\(p:ps) -> zipWith (\ x y -> [x, y])
        (p:ps) (ps++[p])) $ permute is
    poel = [ concat hs |
            uitsteeksels <- extnodes,
            organen <- intnodes,
            let hs = zipWith (:) uitsteeksels organen,
            let subsom = map sum hs,
            length (nub subsom) == 1 ]

Problem 69

Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.

Solution:

{-phi(n) = n*(1-1/p1)*(1-1/p2)*...*(1-1/pn)
n/phi(n) = 1/(1-1/p1)*(1-1/p2)*...*(1-1/pn)
(1-1/p) will be minimal for a small p and 1/(1-1/p) will then be maximal
 -}
primes=[2,3,5,7,11,13,17,19,23]
problem_69=
    maximum [c|
    b<-tail $ inits primes,
    let c=product b,
    c<10^6
    ]

Note: credit for arithmetic functions is due to David Amos.

Problem 70

Investigate values of n for which φ(n) is a permutation of n.

Solution:

import Data.List
import Data.Function
isPerm a b = null $ show a \\ show b
flsqr n x=x<(floor.sqrt.fromInteger) n
pairs n1 = 
    fst . minimumBy (compare `on` fn) $ [(m,pm)|a<-gena,b<-genb,let m=a*b,n>m,let pm=m-a-b+1,isPerm m pm]
    where
    n=fromInteger n1
    gena = dropWhile (flsqr n)$  takeWhile (flsqr (2*n))  primes
    genb = dropWhile (flsqr (n `div` 2))$  takeWhile (flsqr n)  primes
    fn (x,px) = fromIntegral x / (fromIntegral px) 

problem_70= pairs (10^7)