Euler problems/21 to 30
Evaluate the sum of all amicable pairs under 10000.
problem_21 = sum [n | n <- [2..9999], let m = eulerTotient n, m > 1, m < 10000, n == eulerTotient m ]
What is the total of all the name scores in the file of first names?
import Data.List import Data.Char problem_22 = do input <- readFile "names.txt" let names = sort $ read$"["++ input++"]" let scores = zipWith score names [1..] print $ show $ sum $ scores where score w i = (i *) $ sum $ map (\c -> ord c - ord 'A' + 1) w
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
import Data.Array n = 28124 abundant n = eulerTotient n - n > n abunds_array = listArray (1,n) $ map abundant [1..n] abunds = filter (abunds_array !) [1..n] rests x = map (x-) $ takeWhile (<= x `div` 2) abunds isSum = any (abunds_array !) . rests problem_23 = putStrLn $ show $ foldl1 (+) $ filter (not . isSum) [1..n]
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
import Data.List fac 0 = 1 fac n = n * fac (n - 1) perms  _=  perms xs n= x:( perms ( delete x $ xs ) (mod n m)) where m=fac$(length(xs) -1) y=div n m x = xs!!y problem_24 = perms "0123456789" 999999
What is the first term in the Fibonacci sequence to contain 1000 digits?
import Data.List fib x |x==0=0 |x==1=1 |x==2=1 |odd x=(fib (d+1))^2+(fib d)^2 |otherwise=(fib (d+1))^2-(fib (d-1))^2 where d=div x 2 phi=(1+sqrt 5)/2 dig x=floor( (fromInteger x-1) * log 10 /log phi) problem_25 = head[a|a<-[dig num..],(>=limit)$fib a] where num=1000 limit=10^(num-1)
Find the value of d < 1000 for which 1/d contains the longest recurring cycle.
problem_26 = fst $ maximumBy (\a b -> snd a `compare` snd b) [(n,recurringCycle n) | n <- [1..999]] where recurringCycle d = remainders d 10  remainders d 0 rs = 0 remainders d r rs = let r' = r `mod` d in case findIndex (== r') rs of Just i -> i + 1 Nothing -> remainders d (10*r') (r':rs)
Find a quadratic formula that produces the maximum number of primes for consecutive values of n.
The following is written in literate Haskell:
> import Data.List To be sure we get the maximum type checking of the compiler, we switch off the default type > default () Generate a list of primes. It works by filtering out numbers that are divisable by a previously found prime > primes :: [Int] > primes = sieve (2 : [3, 5..]) > where > sieve (p:xs) = p : sieve (filter (\x -> x `mod` p > 0) xs) > isPrime :: Int -> Bool > isPrime x = x `elem` (takeWhile (<= x) primes) The lists of values we are going to try for a and b; b must be a prime, as n² + an + b is equal to b when n = 0 > testRangeA :: [Int] > testRangeA = [-1000 .. 1000] > testRangeB :: [Int] > testRangeB = takeWhile (< 1000) primes The search > bestCoefficients :: (Int, Int, Int) > bestCoefficients = > maximumBy (\(x, _, _) (y, _, _) -> compare x y) $ > [f a b | a <- testRangeA, b <- testRangeB] > where Generate a list of results of the quadratic formula (only the contiguous primes) wrap the result in a triple, together with a and b > f :: Int -> Int -> (Int, Int, Int) > f a b = ( length $ contiguousPrimes a b > , a > , b > ) > contiguousPrimes :: Int -> Int -> [Int] > contiguousPrimes a b = takeWhile isPrime (map (quadratic a b) [0..]) The quadratic formula > quadratic :: Int -> Int -> Int -> Int > quadratic a b n = n * n + a * n + b > problem_27 = > do > let (l, a, b) = bestCoefficients > > putStrLn $ "" > putStrLn $ "Problem Euler 27" > putStrLn $ "" > putStrLn $ "The best quadratic formula found is:" > putStrLn $ " n * n + " ++ show a ++ " * n + " ++ show b > putStrLn $ "" > putStrLn $ "The number of primes is: " ++ (show l) > putStrLn $ "" > putStrLn $ "The primes are:" > print $ take l $ contiguousPrimes a b > putStrLn $ ""
What is the sum of both diagonals in a 1001 by 1001 spiral?
corners :: Int -> (Int, Int, Int, Int) corners i = (n*n, 1+(n*(2*m)), 2+(n*(2*m-1)), 3+(n*(2*m-2))) where m = (i-1) `div` 2 n = 2*m+1 sumcorners :: Int -> Int sumcorners i = a+b+c+d where (a, b, c, d) = corners i sumdiags :: Int -> Int sumdiags i | even i = error "not a spiral" | i == 3 = s + 1 | otherwise = s + sumdiags (i-2) where s = sumcorners i problem_28 = sumdiags 1001
You can note that from 1 to 3 there's (+2), and such too for 5, 7 and 9, it then goes up to (+4) 4 times, and so on, adding 2 to the number to add for each level of the spiral. You can so avoid all need for multiplications and just do additions with the following code :
problem_28 = sum . scanl (+) 1 . concatMap (replicate 4) $ [2,4..1000]
How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
problem_29 = length . group . sort $ [a^b | a <- [2..100], b <- [2..100]]
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
import Data.Char limit = snd $ head $ dropWhile (\(a,b) -> a > b) $ zip (map (9^5*) [1..]) (map (10^) [1..]) fifth n = foldr (\a b -> (toInteger(ord a) - 48)^5 + b) 0 $ show n problem_30 = sum $ filter (\n -> n == fifth n) [2..limit]