# Euler problems/21 to 30

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CaleGibbard (Talk | contribs) (remove comment which didn't apply to the solution given. If you want to improve the solutions, do what HenryLaxen is doing, and actually improve them.) |
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Evaluate the sum of all amicable pairs under 10000. |
Evaluate the sum of all amicable pairs under 10000. |
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+ | Solution: |
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+ | (http://www.research.att.com/~njas/sequences/A063990) |
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+ | |||

+ | This is a little slow because of the naive method used to compute the divisors. |
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<haskell> |
<haskell> |
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− | --http://www.research.att.com/~njas/sequences/A063990 |
+ | problem_21 = sum [m+n | m <- [2..9999], let n = divisorsSum ! m, amicable m n] |

− | problem_21 = sum [220, 284, 1184, 1210, 2620, 2924, 5020, 5564, 6232, 6368] |
+ | where amicable m n = m < n && n < 10000 && divisorsSum ! n == m |

+ | divisorsSum = array (1,9999) |
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+ | [(i, sum (divisors i)) | i <- [1..9999]] |
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+ | divisors n = [j | j <- [1..n `div` 2], n `mod` j == 0] |
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+ | </haskell> |
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+ | |||

+ | Here is an alternative using a faster way of computing the sum of divisors. |
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+ | <haskell> |
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+ | problem_21_v2 = sum [n | n <- [2..9999], let m = d n, |
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+ | m > 1, m < 10000, n == d m] |
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+ | d n = product [(p * product g - 1) `div` (p - 1) | |
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+ | g <- group $ primeFactors n, let p = head g |
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+ | ] - n |
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+ | primeFactors = pf primes |
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+ | where |
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+ | pf ps@(p:ps') n |
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+ | | p * p > n = [n] |
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+ | | r == 0 = p : pf ps q |
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+ | | otherwise = pf ps' n |
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+ | where (q, r) = n `divMod` p |
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+ | primes = 2 : filter (null . tail . primeFactors) [3,5..] |
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</haskell> |
</haskell> |
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## Revision as of 17:09, 25 February 2008

## Contents |

## 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 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..]

## 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 = 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

## 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 = putStrLn . show . foldl1 (+) . 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

## 5 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 )

## 6 Problem 26

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

Solution:

problem_26 = head [a | a<-[999,997..], and [isPrime a, isPrime $ a `div` 2]]

## 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

## 9 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]

## 10 Problem 30

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

Solution:

--http://www.research.att.com/~njas/sequences/A052464 problem_30 = sum [4150, 4151, 54748, 92727, 93084, 194979]

I'm sorry, but I find the solution to problem 30 very unsatisfying. I'm using the Euler problems to learn Haskell, so looking up the answer and adding the terms isn't really that helpful. I would like to present the following as a clearer solution that perhaps gives a little more insight into the problem and programming in Haskell. -- Henry Laxen, Feb 20, 2008

problem_30 = sum $ map listToInt (drop 2 ans) -- we drop 2 because the first two members of the ans are 0 and 1, -- which are considered "trivial" solutions and should not count in the sum where maxFirstDigit = (6*9^5 `div` 10^5) + 1 -- The largest number that can be the sum of fifth powers -- is 6*9^5 = 354294, which has 6 digits listToInt n = foldl (\x y -> 10*x+y) 0 n isSumOfPowers p n = (sum $ map (\x -> x^p) n) == listToInt n ans = filter (isSumOfPowers 5) [ [a,b,c,d,e,f] | a <- [0..maxFirstDigit], b <- [0..9], c <- [0..9], d <- [0..9], e <- [0..9], f <- [0..9] ]