Euler problems/71 to 80
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[edit] 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)
[edit] 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^(i1) * (p1)) factors) where factors = fstfac n fstfac x = [(head a ,length a)a<group$primeFactors x] p72 n= sum [eulerTotient xx < [groups*n+1..groups*(n+1)]] problem_72 = sum [p72 xx < [0..999]]
[edit] 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 20080304
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)
[edit] 4 Problem 74
Determine the number of factorial chains that contain exactly sixty nonrepeating 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:ba<[minnum ..9],b<makeIncreas (digits1) 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[kk<[1..9999],chain k==60]+p74 test = print $ [aa<tail$makeIncreas 6 0,let k=digitToN a,chain k==60]
[edit] 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..n1], 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
[edit] 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
[edit] 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
[edit] 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) [nx x < suite] problem_78 :: Int problem_78 = head $ filter (\x > (partitions ! x) `mod` 1000000 == 0) [1..]
[edit] 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
[edit] 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