# Euler problems/1 to 10

### From HaskellWiki

(→Problem 8: correct the code) |
(→Problem 3) |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | primes = 2 : filter ((==1) . length . primeFactors) [3,5..] |
+ | primes = 2 : filter (null . tail . primeFactors) [3,5..] |

primeFactors n = factor n primes |
primeFactors n = factor n primes |
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| p*p > n = [n] |
| p*p > n = [n] |
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| n `mod` p == 0 = p : factor (n `div` p) (p:ps) |
| n `mod` p == 0 = p : factor (n `div` p) (p:ps) |
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− | | otherwise = factor n ps |
+ | | otherwise = factor n ps |

problem_3 = last (primeFactors 600851475143) |
problem_3 = last (primeFactors 600851475143) |

## Latest revision as of 07:05, 3 September 2014

## Contents |

## [edit] 1 Problem 1

Add all the natural numbers below 1000 that are multiples of 3 or 5.

Two solutions usingimport Data.List (union) problem_1' = sum (union [3,6..999] [5,10..999]) problem_1 = sum [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0]

Another solution which uses algebraic relationships:

problem_1 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999 where sumStep s n = s * sumOnetoN (n `div` s) sumOnetoN n = n * (n+1) `div` 2

## [edit] 2 Problem 2

Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed one million.

Solution:

problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, even x] where fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

The following two solutions use the fact that the even-valued terms in the Fibonacci sequence themselves form a Fibonacci-like sequence that satisfies

problem_2 = sumEvenFibs $ numEvenFibsLessThan 1000000 where sumEvenFibs n = (evenFib n + evenFib (n+1) - 2) `div` 4 evenFib n = round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5 numEvenFibsLessThan n = floor $ (log (fromIntegral n - 0.5) + 0.5*log 5) / log (2 + sqrt 5)

The first two solutions work because 10^6 is small. The following solution also works for much larger numbers (up to at least 10^1000000 on my computer):

problem_2 = sumEvenFibsLessThan 1000000 sumEvenFibsLessThan n = (a + b - 1) `div` 2 where n2 = n `div` 2 (a, b) = foldr f (0,1) . takeWhile ((<= n2) . fst) . iterate times2E $ (1, 4) f x y | fst z <= n2 = z | otherwise = y where z = x `addE` y addE (a, b) (c, d) = (a*d + b*c - 4*ac, ac + b*d) where ac=a*c times2E (a, b) = addE (a, b) (a, b)

## [edit] 3 Problem 3

Find the largest prime factor of 600851475143.

Solution:

primes = 2 : filter (null . tail . primeFactors) [3,5..] primeFactors n = factor n primes where factor n (p:ps) | p*p > n = [n] | n `mod` p == 0 = p : factor (n `div` p) (p:ps) | otherwise = factor n ps problem_3 = last (primeFactors 600851475143)

## [edit] 4 Problem 4

Find the largest palindrome made from the product of two 3-digit numbers.

Solution:

problem_4 = maximum [x | y<-[100..999], z<-[y..999], let x=y*z, let s=show x, s==reverse s]

## [edit] 5 Problem 5

What is the smallest number divisible by each of the numbers 1 to 20?

Solution:

problem_5 = foldr1 lcm [1..20]

Another solution: `16*9*5*7*11*13*17*19`

. Product of maximal powers of primes in the range.

## [edit] 6 Problem 6

What is the difference between the sum of the squares and the square of the sums?

Solution:

problem_6 = (sum [1..100])^2 - sum (map (^2) [1..100])

## [edit] 7 Problem 7

Find the 10001st prime.

Solution:

--primes in problem_3 problem_7 = primes !! 10000

## [edit] 8 Problem 8

Discover the largest product of thirteen consecutive digits in the 1000-digit number.

Solution:

import Data.Char import Data.List euler_8 = do str <- readFile "number.txt" print . maximum . map product . foldr (zipWith (:)) (repeat []) . take 13 . tails . map (fromIntegral . digitToInt) . concat . lines $ str

## [edit] 9 Problem 9

There is only one Pythagorean triplet, {*a*, *b*, *c*}, for which *a* + *b* + *c* = 1000. Find the product *abc*.

Solution:

triplets l = [[a,b,c] | m <- [2..limit], n <- [1..(m-1)], let a = m^2 - n^2, let b = 2*m*n, let c = m^2 + n^2, a+b+c==l] where limit = floor . sqrt . fromIntegral $ l problem_9 = product . head . triplets $ 1000

## [edit] 10 Problem 10

Calculate the sum of all the primes below one million.

Solution:

--primes in problem_3 problem_10 = sum (takeWhile (< 1000000) primes)