# Euler problems/71 to 80

## 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)
```

## 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]]
```

## 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)
```

## 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]
```

## Problem 75

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

Solution:

```module Main where

import Data.Array.Unboxed (UArray, accumArray, elems)

main :: IO ()
main = print problem_75

limit :: Int
limit = 2 * 10 ^ 6

triangs :: [Int]
triangs = [p | n <- [2 .. 1000], m <- [1 .. n - 1], odd (m + n),
m `gcd` n == 1, let p = 2 * (n ^ 2 + m * n), p <= limit]

problem_75 :: Int
problem_75 = length \$ filter (== 1) \$ elems \$
(\ns -> accumArray (+) 0 (1, limit) [(n, 1) | n <- ns] :: UArray Int Int) \$
take limit \$ concatMap (\m -> takeWhile (<= limit) [m, 2 * m .. ]) triangs
```

## 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
```

## 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
```

## 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..]
```

## 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
print \$p79 f
```

## 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
```