# Euler problems/1 to 10

### From HaskellWiki

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<haskell> | <haskell> | ||

− | --http://www.research.att.com/~njas/sequences/A003418 | + | --http://www.research.att.com/~njas/sequences/A003418 (broken URL) |

problem_5 = foldr1 lcm [1..20] | problem_5 = foldr1 lcm [1..20] | ||

</haskell> | </haskell> |

## Revision as of 16:42, 4 September 2012

## Contents |

## 1 Problem 1

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

Two solutions using sum:

import 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

## 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_v2 = sumEvenFibs $ numEvenFibsLessThan 1000000 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)

## 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

## 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 ]

## 5 Problem 5

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

Solution:

--http://www.research.att.com/~njas/sequences/A003418 (broken URL) problem_5 = foldr1 lcm [1..20]

## 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])

## 7 Problem 7

Find the 10001st prime.

Solution:

--primes in problem_3 problem_7 = primes !! 10000

## 8 Problem 8

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

Solution:

import Data.Char groupsOf _ [] = [] groupsOf n xs = take n xs : groupsOf n ( tail xs ) problem_8 x = maximum . map product . groupsOf 5 $ x main = do t <- readFile "p8.log" let digits = map digitToInt $concat $ lines t print $ problem_8 digits

## 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

## 10 Problem 10

Calculate the sum of all the primes below one million.

Solution:

--http://www.research.att.com/~njas/sequences/A046731 problem_10 = sum (takeWhile (< 1000000) primes)