# Euler problems/21 to 30

(Difference between revisions)

## 1 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
[(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..]```

## 2 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 =
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```

## 3 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 . sum . filter (not . isSum) \$ [1..n]```

## 4 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```

Or, using Data.List.permutations,

```import Data.List
problem_24 = (!! 999999) . sort \$ permutations ['0'..'9']```

Casey Hawthorne

For Project Euler #24 you don't need to generate all the lexicographic permutations by Knuth's method or any other.

You're only looking for the millionth lexicographic permutation of "0123456789"

```-- Plan of attack.

-- The "x"s are different numbers
-- 0xxxxxxxxx represents 9! = 362880 permutations/numbers
-- 1xxxxxxxxx represents 9! = 362880 permutations/numbers
-- 2xxxxxxxxx represents 9! = 362880 permutations/numbers

-- 20xxxxxxxx represents 8! = 40320
-- 21xxxxxxxx represents 8! = 40320

-- 23xxxxxxxx represents 8! = 40320
-- 24xxxxxxxx represents 8! = 40320
-- 25xxxxxxxx represents 8! = 40320
-- 26xxxxxxxx represents 8! = 40320
-- 27xxxxxxxx represents 8! = 40320

module Euler where

import Data.List

factorial n = product [1..n]

-- lexOrder "0123456789" 1000000 ""

lexOrder digits left s
| len == 0              = s ++ digits
| quot > 0 && rem == 0  = lexOrder (digits\\(show (digits!!(quot-1))))  rem (s ++ [(digits!!(quot-1))])
| quot == 0 && rem == 0 = lexOrder (digits\\(show (digits!!len)))       rem (s ++ [(digits!!len)])
| rem == 0              = lexOrder (digits\\(show (digits!!(quot+1))))  rem (s ++ [(digits!!(quot+1))])
| otherwise             = lexOrder (digits\\(show (digits!!(quot))))    rem (s ++ [(digits!!(quot))])
where
len = (length digits) - 1
(quot,rem) = quotRem left (factorial len)```

## 5 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```

Casey Hawthorne

I believe you mean the following:

```fibs = 0:1:(zipWith (+) fibs (tail fibs))

last (takeWhile (<10^1000) fibs)```

## 6 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)```

## 7 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```

## 8 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`
Alternatively, one can use the fact that the distance between the diagonal numbers increases by 2 in every concentric square. Each square contains four gaps, so the following
scanl
does the trick:
```euler28 n = sum \$ scanl (+) 0
(1:(concatMap (replicate 4) [2,4..(n-1)]))```

## 9 Problem 29

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

Solution:

```import Control.Monad
problem_29 = length . group . sort \$ liftM2 (^) [2..100] [2..100]```

We can also solve it in a more naive way, without using Monads, like this:

```import List
problem_29 = length \$ nub pr29_help
where pr29_help  = [z | y <- [2..100],
z <- lift y]
lift y = map (\x -> x^y) [2..100]```

Simpler:

```import List
problem_29 = length \$ nub [x^y | x <- [2..100], y <- [2..100]]```

Instead of using lists, the Set data structure can be used for a significant speed increase:

```import Set
problem_29 = size \$ fromList [x^y | x <- [2..100], y <- [2..100]]```

## 10 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 = sum . map ((^5) . toInteger . digitToInt) . show

problem_30 :: Integer
problem_30 = sum \$ filter (\n -> n == fifth n) [2..limit]```