# Euler problems/71 to 80

### From HaskellWiki

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Solution: |
Solution: |
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+ | Brute force but still finds the solution in less than one second. |
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<haskell> |
<haskell> |
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− | problem_77 = undefined |
+ | combinations acc 0 _ = [acc] |

+ | combinations acc _ [] = [] |
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+ | combinations acc value prim@(x:xs) = combinations (acc ++ [x]) value' prim' ++ combinations acc value xs |
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+ | where |
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+ | value' = value - x |
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+ | prim' = dropWhile (>value') prim |
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+ | |||

+ | problem_77 :: Integer |
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+ | problem_77 = head $ filter f [1..] |
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+ | where |
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+ | f n = (length $ combinations [] n $ takeWhile (<n) primes) > 5000 |
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</haskell> |
</haskell> |
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## Revision as of 16:26, 22 August 2007

## Contents |

## 1 Problem 71

Listing reduced proper fractions in ascending order of size.

Solution:

import Data.Ratio (Ratio, (%), numerator) fractions :: [Ratio Integer] fractions = [f | d <- [1..1000000], let n = (d * 3) `div` 7, let f = n%d, f /= 3%7] problem_71 :: Integer problem_71 = numerator $ maximum $ fractions

## 2 Problem 72

How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000?

Solution:

Using the Farey Sequence method, the solution is the sum of phi (n) from 1 to 1000000.

See problem 69 for phi function

problem_72 = sum [phi x|x <- [1..1000000]]

## 3 Problem 73

How many fractions lie between 1/3 and 1/2 in a sorted set of reduced proper fractions?

Solution:

import Data.Ratio (Ratio, (%), numerator, denominator) median :: Ratio Int -> Ratio Int -> Ratio Int median a b = ((numerator a) + (numerator b)) % ((denominator a) + (denominator b)) count :: Ratio Int -> Ratio Int -> Int count a b | d > 10000 = 1 | otherwise = count a m + count m b where m = median a b d = denominator m problem_73 :: Int problem_73 = (count (1%3) (1%2)) - 1

## 4 Problem 74

Determine the number of factorial chains that contain exactly sixty non-repeating terms.

Solution:

import Data.Array (Array, array, (!), elems) import Data.Char (ord) import Data.List (foldl1') import Prelude hiding (cycle) fact :: Integer -> Integer fact 0 = 1 fact n = foldl1' (*) [1..n] factorDigits :: Array Integer Integer factorDigits = array (0,2177281) [(x,n)|x <- [0..2177281], let n = sum $ map (\y -> fact (toInteger $ ord y - 48)) $ show x] cycle :: Integer -> Integer cycle 145 = 1 cycle 169 = 3 cycle 363601 = 3 cycle 1454 = 3 cycle 871 = 2 cycle 45361 = 2 cycle 872 = 2 cycle 45362 = 2 cycle _ = 0 isChainLength :: Integer -> Integer -> Bool isChainLength len n | len < 0 = False | t = isChainLength (len-1) n' | otherwise = (len - c) == 0 where c = cycle n t = c == 0 n' = factorDigits ! n -- | strict version of the maximum function maximum' :: (Ord a) => [a] -> a maximum' [] = undefined maximum' [x] = x maximum' (a:b:xs) = let m = max a b in m `seq` maximum' (m : xs) problem_74 :: Int problem_74 = length $ filter (\(_,b) -> isChainLength 59 b) $ zip ([0..] :: [Integer]) $ take 1000000 $ elems factorDigits

## 5 Problem 75

Find the number of different lengths of wire can that can form a right angle triangle in only one way.

Solution: This is only slightly harder than problem 39. The search condition is simpler but the search space is larger.

problem_75 = length . filter ((== 1) . length) $ group perims where perims = sort [scale*p | p <- pTriples, scale <- [1..10^6 `div` p]] pTriples = [p | n <- [1..1000], m <- [n+1..1000], even n || even m, gcd n m == 1, let a = m^2 - n^2, let b = 2*m*n, let c = m^2 + n^2, let p = a + b + c, p <= 10^6]

## 6 Problem 76

How many different ways can one hundred be written as a sum of at least two positive integers?

Solution:

Calculated using Euler's pentagonal formula and a list for memoization.

partitions = 1 : [sum [s * partitions !! p| (s,p) <- zip signs $ parts n]| n <- [1..]] where signs = cycle [1,1,(-1),(-1)] suite = map penta $ concat [[n,(-n)]|n <- [1..]] penta n = n*(3*n - 1) `div` 2 parts n = takeWhile (>= 0) [n-x| x <- suite] problem_76 = partitions !! 100 - 1

## 7 Problem 77

What is the first value which can be written as the sum of primes in over five thousand different ways?

Solution:

Brute force but still finds the solution in less than one second.

combinations acc 0 _ = [acc] combinations acc _ [] = [] combinations acc value prim@(x:xs) = combinations (acc ++ [x]) value' prim' ++ combinations acc value xs where value' = value - x prim' = dropWhile (>value') prim problem_77 :: Integer problem_77 = head $ filter f [1..] where f n = (length $ combinations [] n $ takeWhile (<n) primes) > 5000

## 8 Problem 78

Investigating the number of ways in which coins can be separated into piles.

Solution:

problem_78 = undefined

## 9 Problem 79

By analysing a user's login attempts, can you determine the secret numeric passcode?

Solution:

problem_79 = undefined

## 10 Problem 80

Calculating the digital sum of the decimal digits of irrational square roots.

Solution:

problem_80 = undefined