# Euler problems/71 to 80

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## Revision as of 07:51, 20 January 2008

## Contents |

## 1 Problem 71

Listing reduced proper fractions in ascending order of size.

Solution:

-- http://mathworld.wolfram.com/FareySequence.html import Data.Ratio ((%), numerator,denominator) fareySeq a b |da2<=10^6=fareySeq a1 b |otherwise=na where na=numerator a nb=numerator b da=denominator a db=denominator b a1=(na+nb)%(da+db) da2=denominator a1 problem_71=fareySeq (0%1) (3%7)

## 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.

groups=1000 eulerTotient n = product (map (\(p,i) -> p^(i-1) * (p-1)) factors) where factors = fstfac n fstfac x = [(head a ,length a)|a<-group$primeFactors x] p72 n= sum [eulerTotient x|x <- [groups*n+1..groups*(n+1)]] problem_72 = sum [p72 x|x <- [0..999]]

## 3 Problem 73

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

Solution:

import Data.Array twix k = crude k - fd2 - sum [ar!(k `div` m) | m <- [3 .. k `div` 5], odd m] where fd2 = crude (k `div` 2) ar = array (5,k `div` 3) $ ((5,1):[(j, crude j - sum [ar!(j `div` m) | m <- [2 .. j `div` 5]]) | j <- [6 .. k `div` 3]]) crude j = m*(3*m+r-2) + s where (m,r) = j `divMod` 6 s = case r of 5 -> 1 _ -> 0 problem_73 = twix 10000

## 4 Problem 74

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

Solution:

import Data.List explode 0 = [] explode n = n `mod` 10 : explode (n `quot` 10) chain 2 = 1 chain 1 = 1 chain 145 = 1 chain 40585 = 1 chain 169 = 3 chain 363601 = 3 chain 1454 = 3 chain 871 = 2 chain 45361 = 2 chain 872 = 2 chain 45362 = 2 chain x = 1 + chain (sumFactDigits x) makeIncreas 1 minnum = [[a]|a<-[minnum..9]] makeIncreas digits minnum = [a:b|a<-[minnum ..9],b<-makeIncreas (digits-1) a] p74= sum[div p6 $countNum a| a<-tail$makeIncreas 6 1, let k=digitToN a, chain k==60 ] where p6=facts!! 6 sumFactDigits = foldl' (\a b -> a + facts !! b) 0 . explode factorial n = if n == 0 then 1 else n * factorial (n - 1) digitToN = foldl' (\a b -> 10*a + b) 0 .dropWhile (==0) facts = scanl (*) 1 [1..9] countNum xs=ys where ys=product$map (factorial.length)$group xs problem_74= length[k|k<-[1..9999],chain k==60]+p74 test = print $ [a|a<-tail$makeIncreas 6 0,let k=digitToN a,chain k==60]

## 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:

Here is a simpler solution: For each n, we create the list of the number of partitions of n whose lowest number is i, for i=1..n. We build up the list of these lists for n=0..100.

build x = (map sum (zipWith drop [0..] x) ++ [1]) : x problem_76 = (sum $ head $ iterate build [] !! 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.

counter = foldl (\without p -> let (poor,rich) = splitAt p without with = poor ++ zipWith (+) with rich in with ) (1 : repeat 0) problem_77 = find ((>5000) . (ways !!)) $ [1..] where ways = counter $ take 100 primes

## 8 Problem 78

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

Solution:

Same as problem 76 but using array instead of lists to speedup things.

import Data.Array partitions :: Array Int Integer partitions = array (0,1000000) $ (0,1) : [(n,sum [s * partitions ! p| (s,p) <- zip signs $ parts n])| n <- [1..1000000]] 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_78 :: Int problem_78 = head $ filter (\x -> (partitions ! x) `mod` 1000000 == 0) [1..]

## 9 Problem 79

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

Solution:

A bit ugly but works fine

import Data.List problem_79 :: String -> String problem_79 file = map fst $ sortBy (\(_,a) (_,b) -> compare (length b) (length a)) $ zip digs order where nums = lines file digs = map head $ group $ sort $ filter (\c -> c >= '0' && c <= '9') file prec = concatMap (\(x:y:z:_) -> [[x,y],[y,z],[x,z]]) nums order = map (\n -> map head $ group $ sort $ map (\(_:x:_) -> x) $ filter (\(x:_) -> x == n) prec) digs main=do f<-readFile "keylog.txt" print$problem_79 f

## 10 Problem 80

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

Solution:

import Data.List ((\\)) hundreds :: Integer -> [Integer] hundreds n = hundreds' [] n where hundreds' acc 0 = acc hundreds' acc n = hundreds' (m : acc) d where (d,m) = divMod n 100 squareDigs :: Integer -> [Integer] squareDigs n = p : squareDigs' p r xs where (x:xs) = hundreds n ++ repeat 0 p = floor $ sqrt $ fromInteger x r = x - (p^2) squareDigs' :: Integer -> Integer -> [Integer] -> [Integer] squareDigs' p r (x:xs) = x' : squareDigs' (p*10 + x') r' xs where n = 100*r + x (x',r') = last $ takeWhile (\(_,a) -> a >= 0) $ scanl (\(_,b) (a',b') -> (a',b-b')) (0,n) rs rs = [y|y <- zip [1..] [(20*p+1),(20*p+3)..]] sumDigits n = sum $ take 100 $ squareDigs n problem_80 :: Integer problem_80 = sum $ map sumDigits [x|x <- [1..100] \\ [n^2|n<-[1..10]]]