Euler problems/1 to 10

Problem 1

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

Solution:

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

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,
    x `mod` 2 == 0
    ]
    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 evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1).

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)

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)

Problem 4

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

Solution:

problem_4 = 
    foldr1 max [ x |
    y <- [100..999],
    z <- [y..999],
    let x = y * z,
    let s = show x,
    s == reverse s
    ]

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
problem_5 = foldr1 lcm [1..20]

Problem 6

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

Solution:

fun n=
    a-b
    where
    a=div (n^2 * (n+1)^2) 4
    b=div (n * (n+1) * (2*n+1)) 6
problem_6=fun 100

Problem 7

Find the 10001st prime.

Solution:

--primes in problem_3
problem_7 = 
    head $ drop 10000 primes

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 $foldl (++) "" $ lines t
    print $ problem_8 digits

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

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)