# Euler problems/1 to 10

## Problem 1

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

Solution:

```problem_1 =
sum [ x |
x <- [1..999],
(x `mod` 3 == 0) ||  (x `mod` 5 == 0)
]
```
```problem_1_v2 =
sum \$ filter (\x -> ( x `mod` 3 == 0 || x `mod` 5 == 0 ) ) [1..999]
```

```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) =
```

## 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)
```

This can be improved by using `null . tail` instead of `(== 1) . length`.

## Problem 4

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

Solution:

```problem_4 =
foldr max 0 [ x |
y <- [100..999],
z <- [100..999],
let x = y * z,
let s = show x,
s == reverse s
]
```

An alternative to avoid evaluating twice the same pair of numbers:

```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:

```problem_5 =
x <- [2520,5040..],
all (\y -> x `mod` y == 0) [1..20]
]
```

An alternative solution that takes advantage of the Prelude to avoid use of the generate and test idiom:

```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:

```problem_6 =
sum [ x^2 | x <- [1..100]] - (sum [1..100])^2
```

## Problem 7

Find the 10001st prime.

Solution:

```--primes in problem_3
problem_7 =
```

As above, this can be improved by using `null . tail` instead of `(== 1) . length`.

Here is an alternative that uses a sieve of Eratosthenes:

```primes' =
2 : 3 : sieve (tail primes') [5,7..]
where
sieve (p:ps) x =
h ++ sieve ps (filter (\q -> q `mod` p /= 0) t
where
(h, _:t) = span (p*p <) x
problem_7_v2 = primes' !! 10000
```

## 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
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:

```problem_9 =
a <- [1..500],
b <- [a..500],
let c = 1000-a-b,
a^2 + b^2 == c^2
]
```

Another solution using Pythagorean Triplets generation:

```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:

```problem_10 =
sum (takeWhile (< 1000000) primes)
```