# 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.

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]
sum1to n = n * (n+1) `div` 2

problem_1_v3 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999
where sumStep s n = s * sum1to (n `div` s)

## 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, x `mod` 2 == 0]
where fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

## 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)

## 4 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]

## 5 Problem 5

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

Solution:

problem_5 = head [ x | 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]

## 6 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

## 7 Problem 7

Find the 10001st prime.

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_7 = head \$ drop 10000 primes

## 8 Problem 8

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

Solution:

num = ... -- 1000 digit number as a string
digits = map digitToInt num

groupsOf _ [] = []
groupsOf n xs = take n xs : groupsOf n ( tail xs )

problem_8 = maximum . map product . groupsOf 5 \$ digits

## 9 Problem 9

There is only one Pythagorean triplet, {a, b, c}, for which a + b + c = 1000. Find the product abc.

Solution:

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

## 10 Problem 10

Calculate the sum of all the primes below one million.

Solution:

problem_10 = sum (takeWhile (< 1000000) primes)