# Euler problems/1 to 10

(Difference between revisions)

## 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

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

## 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 (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)```

## 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 (broken URL)
--primes in problem_3
problem_10 = sum (takeWhile (< 1000000) primes)```