# Euler problems/111 to 120

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## Latest revision as of 08:07, 23 February 2010

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## [edit] 1 Problem 111

Search for 10-digit primes containing the maximum number of repeated digits.

Solution:

import Control.Monad (replicateM) -- All ways of interspersing n copies of x into a list intr :: Int -> a -> [a] -> [[a]] intr 0 _ y = [y] intr n x (y:ys) = concat [map ((replicate i x ++) . (y :)) $ intr (n-i) x ys | i <- [0..n]] intr n x _ = [replicate n x] -- All 10-digit primes containing the maximal number of the digit d maxDigits :: Char -> [Integer] maxDigits d = head $ dropWhile null [filter isPrime $ map read $ filter ((/='0') . head) $ concatMap (intr (10-n) d) $ replicateM n $ delete d "0123456789" | n <- [1..9]] problem_111 = sum $ concatMap maxDigits "0123456789"

## [edit] 2 Problem 112

Investigating the density of "bouncy" numbers.

Solution:

import Data.List isIncreasing x = show x == sort (show x) isDecreasing x = reverse (show x) == sort (show x) isBouncy x = not (isIncreasing x) && not (isDecreasing x) findProportion prop = snd . head . filter condition . zip [1..] where condition (a,b) = a >= prop * fromIntegral b problem_112 = findProportion 0.99 $ filter isBouncy [1..]

## [edit] 3 Problem 113

How many numbers below a googol (10100) are not "bouncy"?

Solution:

import Array mkArray b f = listArray b $ map f (range b) digits = 100 inc = mkArray ((1, 0), (digits, 9)) ninc dec = mkArray ((1, 0), (digits, 9)) ndec ninc (1, _) = 1 ninc (l, d) = sum [inc ! (l-1, i) | i <- [d..9]] ndec (1, _) = 1 ndec (l, d) = sum [dec ! (l-1, i) | i <- [0..d]] problem_113 = sum [inc ! i | i <- range ((digits, 0), (digits, 9))] + sum [dec ! i | i <- range ((1, 1), (digits, 9))] - digits*9 -- numbers like 11111 are counted in both inc and dec - 1 -- 0 is included in the increasing numbers

Note: inc and dec contain the same data, but it seems clearer to duplicate them.

it is another way to solution this problem:

binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y)) prodxy x y=product[x..y] problem_113=sum[binomial (8+a) a+binomial (9+a) a-10|a<-[1..100]]

## [edit] 4 Problem 114

Investigating the number of ways to fill a row with separated blocks that are at least three units long.

Solution:

-- fun in p115 problem_114=fun 3 50

## [edit] 5 Problem 115

Finding a generalisation for the number of ways to fill a row with separated blocks.

Solution:

binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y)) prodxy x y=product[x..y] fun m n=sum[binomial (k+a) (k-a)|a<-[0..div (n+1) (m+1)],let k=1-a*m+n] problem_115 = (+1)$length$takeWhile (<10^6) [fun 50 i|i<-[1..]]

## [edit] 6 Problem 116

Investigating the number of ways of replacing square tiles with one of three coloured tiles.

Solution:

binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y)) prodxy x y=product[x..y] f116 n x=sum[binomial (a+b) a|a<-[1..div n x],let b=n-a*x] p116 x=sum[f116 x a|a<-[2..4]] problem_116 = p116 50

## [edit] 7 Problem 117

Investigating the number of ways of tiling a row using different-sized tiles.

Solution:

fibs5 = 0 : 0 :1: 1:zipWith4 (\a b c d->a+b+c+d) fibs5 a1 a2 a3 where a1=tail fibs5 a2=tail a1 a3=tail a2 p117 x=fibs5!!(x+2) problem_117 = p117 50

## [edit] 8 Problem 118

Exploring the number of ways in which sets containing prime elements can be made.

Solution:

digits = ['1'..'9'] -- possible partitions voor prime number sets -- leave out patitions with more than 4 1's -- because only {2,3,5,7,..} is possible -- and the [9]-partition because every permutation of all -- nine digits is divisable by 3 test xs |len>4=False |xs==[9]=False |otherwise=True where len=length $filter (==1) xs parts = filter test $partitions 9 permutationsOf [] = [[]] permutationsOf xs = [x:xs' | x <- xs, xs' <- permutationsOf (delete x xs)] combinationsOf 0 _ = [[]] combinationsOf _ [] = [] combinationsOf k (x:xs) = map (x:) (combinationsOf (k-1) xs) ++ combinationsOf k xs priemPerms [] = 0 priemPerms ds = fromIntegral . length . filter (isPrime . read) . permutationsOf $ ds setsums [] 0 = [[]] setsums [] _ = [] setsums (x:xs) n | x > n = setsums xs n | otherwise = map (x:) (setsums (x:xs) (n-x)) ++ setsums xs n partitions n = setsums (reverse [1..n]) n fc :: [Integer] -> [Char] -> Integer fc (p:[]) ds = priemPerms ds fc (p:ps) ds = sum [np y * fc ps (ds \\ y) | y <- combinationsOf p ds, np y /= 0] where np = priemPerms -- here is the 'imperfection' correction method: -- make use of duplicate reducing factors for partitions -- with repeating factors, f.i. [1,1,1,1,2,3]: -- in this case 4 1's -> factor = 4! -- or for [1,1,1,3,3] : factor = 3! * 2! dupF :: [Integer] -> Integer dupF = product . map (product . enumFromTo 1 . fromIntegral . length) . group main = do print . sum . map (\x -> fc x digits `div` dupF x) $ parts problem_118 = main

## [edit] 9 Problem 119

Investigating the numbers which are equal to sum of their digits raised to some power.

Solution:

import Data.List digits n {- 123->[3,2,1] -} |n<10=[n] |otherwise= y:digits x where (x,y)=divMod n 10 problem_119 =sort [(a^b)| a<-[2..200], b<-[2..9], let m=a^b, let n=sum$digits m, n==a]!!29

## [edit] 10 Problem 120

Finding the maximum remainder when (a − 1)^{n} + (a + 1)^{n} is divided by a^{2}.

Solution:

fun m=div (m*(8*m^2-3*m-5)) 3 problem_120 = fun 500

I have no idea what the above solution has to do with this
problem, even though it produces the correct answer. I suspect
it is some kind of red herring. Below you will find a more holy
mackerel approach, based on the observation that:

1. (a-1)^{n} + (a+1)^{n} = 2 if n is odd, and 2an if n is even (mod a^{2})

2. the maximum of 2an mod a^{2} occurs when n = (a-1)/2

I hope this is a little more transparent than the solution proposed above. Henrylaxen Mar 5, 2008

maxRemainder n = 2 * n * ((n-1) `div` 2) problem_120 = sum $ map maxRemainder [3..1000]