# Euler problems/91 to 100

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## 1 Problem 91

Find the number of right angle triangles in the quadrant.

Solution:

```reduce x y = (quot x d, quot y d)
where d = gcd x y

problem_91 n = 3*n*n + 2* sum others
where
others = do
x1 <- [1..n]
y1 <- [1..n]
let (yi,xi) = reduce x1 y1
let yc = quot (n-y1) yi
let xc = quot x1 xi
return (min xc yc)```

## 2 Problem 92

Investigating a square digits number chain with a surprising property.

Solution:

```import Data.List
import Data.Char
pow2=[a*a|a<-[0..9]]
sum_pow 1=0
sum_pow 89=1
sum_pow x= sum_pow(sum [pow2 !! y |y<-digits x])
digits n
{-  change 123 to [3,2,1]
-}
|n<10=[n]
|otherwise= y:digits x
where
(x,y)=divMod n 10
groups=1000
problem_92 b=sum [sum_pow a|a<-[1+b*groups..groups+b*groups]]
gogo li
=if (li>9999)
then return()
else do appendFile "file.log" ((show\$problem_92 li)  ++ "\n")
gogo (li+1)
main=gogo 0```

## 3 Problem 93

Using four distinct digits and the rules of arithmetic, find the longest sequence of target numbers.

Solution:

`problem_93 = undefined`

## 4 Problem 94

Investigating almost equilateral triangles with integral sides and area.

Solution:

`problem_94 = undefined`

## 5 Problem 95

Find the smallest member of the longest amicable chain with no element exceeding one million.

Solution which avoid visiting a number more than one time :

```import Data.Array.Unboxed
import qualified Data.IntSet as S
import Data.List

takeUntil _ [] = []
takeUntil pred (x:xs) = x : if pred x then takeUntil pred xs else []

chain n s =  lgo [n] \$ properDivisorsSum ! n
where lgo xs x | x > 1000000 || S.notMember x s = (xs,[])
| x `elem` xs = (xs,x : takeUntil (/= x) xs)
| otherwise = lgo (x:xs) \$ properDivisorsSum ! x

properDivisorsSum :: UArray Int Int
properDivisorsSum = accumArray (+) 1 (0,1000000)
\$ (0,-1):[(k,factor)|
factor<-[2..1000000 `div` 2]
, k<-[2*factor,2*factor+factor..1000000]
]

base = S.fromList [1..1000000]

problem_95 = fst \$ until (S.null . snd) f ((0,0),base)
where
f (p@(n,m), s) = (p', s')
where
setMin = head \$ S.toAscList s
(explored, chn) = chain setMin s
len = length chn
p' = if len > m then (minimum chn, len) else p
s' = foldl' (flip S.delete) s explored```

Here is a more straightforward solution, without optimization. Yet it solves the problem in a few seconds when compiled with GHC 6.6.1 with the -O2 flag. I like to let the compiler do the optimization, without cluttering my code.

This solution avoids using unboxed arrays, which many consider to be somewhat of an imperitive-style hack. In fact, no memoization at all is required.

```import Data.List (foldl1', group)

-- The sum of all proper divisors of n.
d n = product [(p * product g - 1) `div` (p - 1) |
g <- group \$ primeFactors n, let p = head g
] - n

primeFactors = pf primes
where
pf ps@(p:ps') n
| p * p > n = [n]
| r == 0    = p : pf ps q
| otherwise = pf ps' n
where
(q, r) = n `divMod` p

primes = 2 : filter (null . tail . primeFactors) [3,5..]

-- The longest chain of numbers is (n, k), where
-- n is the smallest number in the chain, and k is the length
-- of the chain. We limit the search to chains whose
-- smallest number is no more than m and, optionally, whose
-- largest number is no more than m'.
longestChain m m' = (n, k)
where
(n, Just k) = foldl1' cmpChain [(n, findChain n) | n <- [2..m]]
findChain n = f [] n \$ d n
f s n n'
| n' == n               = Just \$ 1 + length s
| n' < n                = Nothing
| maybe False (< n') m' = Nothing
| n' `elem` s           = Nothing
| otherwise             = f (n' : s) n \$ d n'
cmpChain p@(n, k) q@(n', k')
| (k, negate n) < (k', negate n') = q
| otherwise                       = p

problem_95_v2 = longestChain 1000000 (Just 1000000)```

## 6 Problem 96

Devise an algorithm for solving Su Doku puzzles.

See numerous solutions on the Sudoku page.

## 7 Problem 97

Find the last ten digits of the non-Mersenne prime: 28433 × 27830457 + 1.

Solution:

`problem_97 = (28433 * 2^7830457 + 1) `mod` (10^10)`

## 8 Problem 98

Investigating words, and their anagrams, which can represent square numbers.

Solution:

```import Data.List
import Data.Maybe

-- Replace each letter of a word, or digit of a number, with
-- the index of where that letter or digit first appears
profile :: Ord a => [a] -> [Int]
profile x = map (fromJust . flip lookup (indices x)) x
where
indices = map head . groupBy fstEq . sort . flip zip [0..]

-- Check for equality on the first component of a tuple
fstEq :: Eq a => (a, b) -> (a, b) -> Bool
fstEq x y = (fst x) == (fst y)

-- The histogram of a small list
hist :: Ord a => [a] -> [(a, Int)]
hist = let item g = (head g, length g) in map item . group . sort

-- The list of anagram sets for a word list.
anagrams :: Ord a => [[a]] -> [[[a]]]
anagrams x = map (map snd) \$ filter (not . null . drop 1) \$
groupBy fstEq \$ sort \$ zip (map hist x) x

-- Given two finite lists that are a permutation of one
-- another, return the permutation function
mkPermute :: Ord a => [a] -> [a] -> ([b] -> [b])
mkPermute x y = pairsToPermute \$ concat \$
zipWith zip (occurs x) (occurs y)
where
pairsToPermute ps = flip map (map snd \$ sort ps) . (!!)
occurs = map (map snd) . groupBy fstEq . sort . flip zip [0..]

problem_98 :: [String] -> Int
problem_98 ws = read \$ head
[y | was <- sortBy longFirst \$ anagrams ws,     -- word anagram sets
w1:t <- tails was, w2 <- t,
let permute = mkPermute w1 w2,
nas <- sortBy longFirst \$ anagrams \$
filter ((== profile w1) . profile) \$
dropWhile (flip longerThan w1) \$
takeWhile (not . longerThan w1) \$
map show \$ map (\x -> x * x) [1..], -- number anagram sets
x:t <- tails nas, y <- t,
permute x == y || permute y == x
]

run_problem_98 :: IO Int
run_problem_98 = do
words_file <- readFile "words.txt"
let words = read \$ '[' : words_file ++ "]"
return \$ problem_98 words

-- Sort on length of first element, from longest to shortest
longFirst :: [[a]] -> [[a]] -> Ordering
longFirst (x:_) (y:_) = compareLen y x

-- Is y longer than x?
longerThan :: [a] -> [a] -> Bool
longerThan x y = compareLen x y == LT

-- Compare the lengths of lists, with short-circuiting
compareLen :: [a] -> [a] -> Ordering
compareLen (_:xs) y  = case y of (_:ys) -> compareLen xs ys
_      -> GT
compareLen _      [] = EQ
compareLen _      _  = LT```

## 9 Problem 99

Which base/exponent pair in the file has the greatest numerical value?

Solution:

`problem_99 = undefined`

## 10 Problem 100

Finding the number of blue discs for which there is 50% chance of taking two blue.

Solution:

`problem_100 = undefined`