# Difference between revisions of "Euler problems/21 to 30"

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isSum = any (abunds_array !) . rests |
isSum = any (abunds_array !) . rests |
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− | problem_23 = |
+ | problem_23 = print . foldl1 (+) . filter (not . isSum) $ [1..n] |

</haskell> |
</haskell> |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | problem_26 = fst $ maximumBy ( |
+ | problem_26 = fst $ maximumBy (comparing snd) |

[(n,recurringCycle n) | n <- [1..999]] |
[(n,recurringCycle n) | n <- [1..999]] |
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where recurringCycle d = remainders d 10 [] |
where recurringCycle d = remainders d 10 [] |
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remainders d 0 rs = 0 |
remainders d 0 rs = 0 |
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remainders d r rs = let r' = r `mod` d |
remainders d r rs = let r' = r `mod` d |
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− | in case |
+ | in case elemIndex r' rs of |

Just i -> i + 1 |
Just i -> i + 1 |
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Nothing -> remainders d (10*r') (r':rs) |
Nothing -> remainders d (10*r') (r':rs) |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | import Data.Char ( |
+ | import Data.Char (digitToInt) |

limit :: Integer |
limit :: Integer |
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fifth :: Integer -> Integer |
fifth :: Integer -> Integer |
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− | fifth n = foldr (\a b -> |
+ | fifth n = foldr (\a b -> toInteger(digitToInt a)^5 + b) 0 $ show n |

problem_30 :: Integer |
problem_30 :: Integer |

## Revision as of 07:05, 13 December 2009

## Contents

## Problem 21

Evaluate the sum of all amicable pairs under 10000.

Solution: (http://www.research.att.com/~njas/sequences/A063990)

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 m /= d (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:

```
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 . sum $ scores
where score w i = (i *) . sum . map (\c -> ord c - ord 'A' + 1) $ w
```

## Problem 23

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

Solution:

```
--http://www.research.att.com/~njas/sequences/A048242
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 = print . foldl1 (+) . filter (not . isSum) $ [1..n]
```

## Problem 24

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

Solution:

```
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
```

## Problem 25

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

Solution:

```
fibs = 0:1:(zipWith (+) fibs (tail fibs))
t = 10^999
problem_25 = length w
where
w = takeWhile (< t) fibs
```

## Problem 26

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

Solution:

```
problem_26 = fst $ maximumBy (comparing snd)
[(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 elemIndex 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:

```
problem_27 = -(2*a-1)*(a^2-a+41)
where n = 1000
m = head $ filter (\x->x^2-x+41>n) [1..]
a = m-1
```

## Problem 28

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

Solution:

```
problem_28 = sum (map (\n -> 4*(n-2)^2+10*(n-1)) [3,5..1001]) + 1
```

## Problem 29

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

Solution:

```
import Control.Monad
problem_29 = length . group . sort $ liftM2 (^) [2..100] [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 (digitToInt)
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(digitToInt a)^5 + b) 0 $ show n
problem_30 :: Integer
problem_30 = sum $ filter (\n -> n == fifth n) [2..limit]
```