# 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:

```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
```

## 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: 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]
```

## 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:

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..]
```

## 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 =
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
print\$problem_79 f
```

## 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]]]
```