# Euler problems/21 to 30

## Contents

## Problem 21

Evaluate the sum of all amicable pairs under 10000.

Solution: This is a little slow because of the naive method used to compute the divisors.

```
problem_21 = sum [m+n | m <- [2..9999], let n = divisorsSum ! m, amicable m n]
where amicable m n = m < n && n < 10000 && divisorsSum ! n == m
divisorsSum = array (1,9999)
[(i, sum (divisors i)) | i <- [1..9999]]
divisors n = [j | j <- [1..n `div` 2], n `mod` j == 0]
```

Here is an alternative using a faster way of computing the sum of divisors.

```
problem_21_v2 = sum [n | n <- [2..9999], let m = d n,
m > 1, m < 10000, n == d m]
d n = product [(p * product g - 1) `div` (p - 1) |
g <- group $ primeFactors n, let p = head g
] - n
primeFactors = pf primes
where
pf ps@(p:ps') n
| p * p > n = [n]
| r == 0 = p : pf ps q
| otherwise = pf ps' n
where (q, r) = n `divMod` p
primes = 2 : filter (null . tail . primeFactors) [3,5..]
```

## Problem 22

What is the total of all the name scores in the file of first names?

Solution:

```
-- apply to a list of names
problem_22 :: [String] -> Int
problem_22 = sum . zipWith (*) [ 1 .. ] . map score
where score = sum . map ( subtract 64 . ord )
```

## Problem 23

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

Solution:

```
import Data.Set hiding (filter, map)
import Data.List (scanl, group)
problem_23 :: Integer
problem_23 = sum [1..28123] - (fold (+) 0 $ abundant_sums $ abundant 28123)
abundant_sums :: [Integer] -> Set Integer
abundant_sums [] = empty
abundant_sums l@(x:xs) = union (fromList [x + a | a <- takeWhile (\y -> y <= 28123 - x) l]) (abundant_sums xs)
abundant :: Integer -> [Integer]
abundant n = [a | a <- [1..n], (sum $ factors a) - a > a]
primes :: [Integer]
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
primeFactors :: Integer -> [Integer]
primeFactors n = factor n primes
where
factor _ [] = []
factor m (p:ps) | p*p > m = [m]
| m `mod` p == 0 = p : factor (m `div` p) (p:ps)
| otherwise = factor m ps
factors :: Integer -> [Integer]
factors = perms . map (tail . scanl (*) 1) . group . primeFactors
where
perms :: (Integral a) => [[a]] -> [a]
perms [] = [1]
perms (x:xs) = perms xs ++ concatMap (\z -> map (*z) $ perms xs) x
```

## Problem 24

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

Solution:

```
perms [] = [[]]
perms xs = do
x <- xs
map ( x: ) ( perms . delete x $ xs )
problem_24 = ( perms "0123456789" ) !! 999999
```

## Problem 25

What is the first term in the Fibonacci sequence to contain 1000 digits?

Solution:

```
valid ( i, n ) = length ( show n ) == 1000
problem_25 = fst . head . filter valid . zip [ 1 .. ] $ fibs
where fibs = 1 : 1 : 2 : zipWith (+) fibs ( tail fibs )
```

## Problem 26

Find the value of d < 1000 for which 1/d contains the longest recurring cycle.

Solution:

```
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)
```

## Problem 27

Find a quadratic formula that produces the maximum number of primes for consecutive values of n.

Solution:

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 $ ""
```

## Problem 28

What is the sum of both diagonals in a 1001 by 1001 spiral?

Solution:

```
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]
```

## Problem 29

How many distinct terms are in the sequence generated by a^{b} for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?

Solution:

```
problem_29 = length . group . sort $ [a^b | a <- [2..100], b <- [2..100]]
```

## Problem 30

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

Solution:

```
import Data.Char (ord)
limit :: Integer
limit = snd $ head $ dropWhile (\(a,b) -> a > b) $ zip (map (9^5*) [1..]) (map (10^) [1..])
fifth :: Integer -> Integer
fifth n = foldr (\a b -> (toInteger(ord a) - 48)^5 + b) 0 $ show n
problem_30 :: Integer
problem_30 = sum $ filter (\n -> n == fifth n) [2..limit]
```