# Euler problems/1 to 10

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− | Do them on your own! |
+ | == [http://projecteuler.net/index.php?section=problems&id=1 Problem 1] == |

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

+ | Two solutions using <hask>sum</hask>: |
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+ | <haskell> |
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+ | import Data.List (union) |
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+ | problem_1' = sum (union [3,6..999] [5,10..999]) |
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+ | |||

+ | problem_1 = sum [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0] |
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+ | </haskell> |
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+ | |||

+ | Another solution which uses algebraic relationships: |
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+ | |||

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

+ | == [http://projecteuler.net/index.php?section=problems&id=2 Problem 2] == |
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+ | Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed one million. |
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+ | |||

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

+ | The following two solutions use the fact that the even-valued terms in |
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+ | the Fibonacci sequence themselves form a Fibonacci-like sequence |
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+ | that satisfies |
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+ | <hask>evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)</hask>. |
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+ | <haskell> |
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+ | problem_2 = sumEvenFibs $ numEvenFibsLessThan 1000000 |
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+ | where |
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+ | sumEvenFibs n = (evenFib n + evenFib (n+1) - 2) `div` 4 |
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+ | evenFib n = round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5 |
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+ | numEvenFibsLessThan n = |
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+ | floor $ (log (fromIntegral n - 0.5) + 0.5*log 5) / log (2 + sqrt 5) |
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+ | </haskell> |
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+ | |||

+ | The first two solutions work because 10^6 is small. |
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+ | The following solution also works for much larger numbers |
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+ | (up to at least 10^1000000 on my computer): |
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+ | <haskell> |
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+ | problem_2 = sumEvenFibsLessThan 1000000 |
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+ | |||

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

+ | times2E (a, b) = addE (a, b) (a, b) |
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+ | </haskell> |
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+ | |||

+ | == [http://projecteuler.net/index.php?section=problems&id=3 Problem 3] == |
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+ | Find the largest prime factor of 317584931803. |
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+ | |||

+ | Solution: |
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+ | <haskell> |
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+ | primes = 2 : filter ((==1) . length . primeFactors) [3,5..] |
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+ | |||

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

+ | problem_3 = last (primeFactors 317584931803) |
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+ | </haskell> |
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+ | |||

+ | Another solution, not using recursion, is: |
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+ | <haskell> |
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+ | problem_3 = (m !! 0) `div` (m !! 1) |
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+ | where |
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+ | m = reverse $ |
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+ | takeWhile (<=n) (scanl1 (*) [ x | x <- 2:[3,5..], (n `mod` x) == 0 ]) |
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+ | n = 600851475143 |
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+ | </haskell> |
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+ | |||

+ | == [http://projecteuler.net/index.php?section=problems&id=4 Problem 4] == |
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+ | Find the largest palindrome made from the product of two 3-digit numbers. |
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+ | |||

+ | Solution: |
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+ | <haskell> |
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+ | problem_4 = |
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+ | maximum [x | y<-[100..999], z<-[y..999], let x=y*z, let s=show x, s==reverse s] |
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+ | </haskell> |
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+ | |||

+ | == [http://projecteuler.net/index.php?section=problems&id=5 Problem 5] == |
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+ | What is the smallest number divisible by each of the numbers 1 to 20? |
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+ | |||

+ | Solution: |
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+ | <haskell> |
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+ | problem_5 = foldr1 lcm [1..20] |
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+ | </haskell> |
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+ | |||

+ | == [http://projecteuler.net/index.php?section=problems&id=6 Problem 6] == |
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+ | What is the difference between the sum of the squares and the square of the sums? |
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+ | |||

+ | Solution: |
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+ | <!-- |
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+ | <haskell> |
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+ | fun n = a - b |
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+ | where |
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+ | a=(n^2 * (n+1)^2) `div` 4 |
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+ | b=(n * (n+1) * (2*n+1)) `div` 6 |
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+ | |||

+ | problem_6 = fun 100 |
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+ | </haskell> |
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+ | --> |
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+ | <!-- Might just be me, but I find this a LOT easier to read. Perhaps not as good mathematically, but it runs in an instant, even for n = 25000. |
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+ | <haskell> |
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+ | fun n = a - b |
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+ | where |
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+ | a = (sum [1..n])^2 |
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+ | b = sum (map (^2) [1..n]) |
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+ | |||

+ | problem_6 = fun 100 |
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+ | </haskell> |
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+ | --> |
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+ | <!-- I just made it a oneliner... --> |
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+ | <haskell> |
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+ | problem_6 = (sum [1..100])^2 - sum (map (^2) [1..100]) |
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+ | </haskell> |
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+ | |||

+ | == [http://projecteuler.net/index.php?section=problems&id=7 Problem 7] == |
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+ | Find the 10001st prime. |
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+ | |||

+ | Solution: |
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+ | <haskell> |
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+ | --primes in problem_3 |
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+ | problem_7 = primes !! 10000 |
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+ | </haskell> |
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+ | == [http://projecteuler.net/index.php?section=problems&id=8 Problem 8] == |
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+ | Discover the largest product of five consecutive digits in the 1000-digit number. |
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+ | |||

+ | Solution: |
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+ | <!-- |
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+ | <haskell> |
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+ | import Data.Char |
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+ | groupsOf _ [] = [] |
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+ | groupsOf n xs = |
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+ | take n xs : groupsOf n ( tail xs ) |
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+ | |||

+ | problem_8 x = maximum . map product . groupsOf 5 $ x |
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+ | main = do t <- readFile "p8.log" |
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+ | let digits = map digitToInt $concat $ lines t |
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+ | print $ problem_8 digits |
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+ | </haskell> |
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+ | --> |
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+ | <!-- I just cleaned up a little. --> |
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+ | <haskell> |
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+ | import Data.Char (digitToInt) |
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+ | import Data.List (tails) |
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+ | |||

+ | problem_8 = do str <- readFile "number.txt" |
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+ | -- This line just converts our str(ing) to a list of 1000 Ints |
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+ | let number = map digitToInt (concat $ lines str) |
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+ | print $ maximum $ map (product . take 5) (tails number) |
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+ | </haskell> |
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+ | |||

+ | == [http://projecteuler.net/index.php?section=problems&id=9 Problem 9] == |
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+ | There is only one Pythagorean triplet, {''a'', ''b'', ''c''}, for which ''a'' + ''b'' + ''c'' = 1000. Find the product ''abc''. |
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+ | |||

+ | Solution: |
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+ | <haskell> |
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+ | triplets l = [[a,b,c] | m <- [2..limit], |
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+ | n <- [1..(m-1)], |
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+ | let a = m^2 - n^2, |
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+ | let b = 2*m*n, |
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+ | let c = m^2 + n^2, |
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+ | a+b+c==l] |
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+ | where limit = floor . sqrt . fromIntegral $ l |
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+ | |||

+ | problem_9 = product . head . triplets $ 1000 |
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+ | </haskell> |
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+ | |||

+ | == [http://projecteuler.net/index.php?section=problems&id=10 Problem 10] == |
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+ | Calculate the sum of all the primes below one million. |
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+ | |||

+ | Solution: |
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+ | <haskell> |
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+ | --primes in problem_3 |
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+ | problem_10 = sum (takeWhile (< 1000000) primes) |
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+ | </haskell> |

## Latest revision as of 14:17, 22 October 2012

## 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 317584931803.

Solution:

primes = 2 : filter ((==1) . length . 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 317584931803)

Another solution, not using recursion, is:

problem_3 = (m !! 0) `div` (m !! 1) where m = reverse $ takeWhile (<=n) (scanl1 (*) [ x | x <- 2:[3,5..], (n `mod` x) == 0 ]) n = 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]

## [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 five consecutive digits in the 1000-digit number.

Solution:

import Data.Char (digitToInt) import Data.List (tails) problem_8 = do str <- readFile "number.txt" -- This line just converts our str(ing) to a list of 1000 Ints let number = map digitToInt (concat $ lines str) print $ maximum $ map (product . take 5) (tails number)

## [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)