# Difference between revisions of "Euler problems/1 to 10"

Robinrobin (talk | contribs) m (Oh come on!) |
CaleGibbard (talk | contribs) (Remove useless comment. The solution doesn't involve a specific definition of the list of primes, so it's not clear how fast or slowly it will run, or whether it is implemented with a sieving method.) |
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Solution: |
Solution: |
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− | Horrible slow of course. You might better learn to make a primesieve. |
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<haskell> |
<haskell> |
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--http://www.research.att.com/~njas/sequences/A046731 |
--http://www.research.att.com/~njas/sequences/A046731 |

## Revision as of 22:21, 24 February 2008

## Contents

## 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 = maximum [ 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 = primes !! 10001
```

## 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)
```