# Euler problems/71 to 80

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(replaced solution to problem 80 with one that a) is complete, b) works, c) doesn't look dreadful) |
m |

## Revision as of 10:37, 13 December 2009

## 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:

If you haven't done so already, read about Farey sequences in Wikipedia http://en.wikipedia.org/wiki/Farey_sequence, where you will learn about mediants. Then divide and conquer. The number of Farey ratios between (a, b) is 1 + the number between (a, mediant a b) + the number between (mediant a b, b). Henrylaxen 2008-03-04

import Data.Ratio mediant :: (Integral a) => Ratio a -> Ratio a -> Ratio a mediant f1 f2 = (numerator f1 + numerator f2) % (denominator f1 + denominator f2) fareyCount :: (Integral a, Num t) => a -> (Ratio a, Ratio a) -> t fareyCount n (a,b) = let c = mediant a b in if (denominator c > n) then 0 else 1 + (fareyCount n (a,c)) + (fareyCount n (c,b)) problem_73 :: Integer problem_73 = fareyCount 10000 (1%3,1%2)

## 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:

import Data.Array triangs :: [Int] triangs = [p | n <- [2..1000], m <- [1..n-1], gcd m n == 1, odd (m+n), let p = 2 * (n^2 + m*n), p <= 2*10^6] problem_75 :: Int problem_75 = length $ filter (\(_, c) -> c == 1) $ assocs $ (\ns -> accumArray (+) 0 (1, 2*10^6) [(n, 1) | n <- ns, inRange (1, 2*10^6) n]) $ concatMap (\n -> takeWhile (<=2*10^6) [n,2*n..]) triangs

## 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:

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:

import Data.Char (digitToInt, intToDigit) import Data.Graph (buildG, topSort) import Data.List (intersect) p79 file= (+0)$read . intersect graphWalk $ usedDigits where usedDigits = intersect "0123456789" $ file edges = concatMap (edgePair . map digitToInt) . words $ file graphWalk = map intToDigit . topSort . buildG (0, 9) $ edges edgePair [x, y, z] = [(x, y), (y, z)] edgePair _ = undefined problem_79 = do f<-readFile "keylog.txt" print $p79 f

## 10 Problem 80

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

This solution uses binary search to find the square root of a large Integer:

import Data.Char (digitToInt) intSqrt :: Integer -> Integer intSqrt n = bsearch 1 n where bsearch l u = let m = (l+u) `div` 2 m2 = m^2 in if u <= l then m else if m2 < n then bsearch (m+1) u else bsearch l m problem_80 :: Int problem_80 = sum [f r | a <- [1..100], let x = a * e, let r = intSqrt x, r*r /= x] where e = 10^202 f = sum . take 100 . map digitToInt . show