# Euler problems/11 to 20

### From HaskellWiki

BrettGiles (Talk | contribs) m (EulerProblems/11 to 20 moved to Euler problems/11 to 20) |
m (added solution 17) |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | problem_17 = undefined |
+ | -- not a very concise or beautiful solution, but food for improvements :) |

+ | |||

+ | names = concat $ |
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+ | [zip [(0, n) | n <- [0..19]] |
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+ | ["", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight" |
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+ | ,"Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen" |
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+ | ,"Sixteen", "Seventeen", "Eighteen", "Nineteen"] |
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+ | ,zip [(1, n) | n <- [0..9]] |
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+ | ["", "Ten", "Twenty", "Thirty", "Fourty", "Fifty", "Sixty", "Seventy" |
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+ | ,"Eighty", "Ninety"] |
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+ | ,[((2,0), "")] |
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+ | ,[((2, n), look (0,n) ++ " Hundred and") | n <- [1..9]] |
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+ | ,[((3,0), "")] |
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+ | ,[((3, n), look (0,n) ++ " Thousand") | n <- [1..9]]] |
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+ | |||

+ | look n = fromJust . lookup n $ names |
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+ | |||

+ | spell n = unwords $ if last s == "and" then init s else s |
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+ | where |
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+ | s = words . unwords $ map look digs' |
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+ | digs = reverse . zip [0..] . reverse . map digitToInt . show $ n |
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+ | digs' = case lookup 1 digs of |
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+ | Just 1 -> |
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+ | let [ten,one] = filter (\(a,_) -> a<=1) digs in |
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+ | (digs \\ [ten,one]) ++ [(0,(snd ten)*10+(snd one))] |
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+ | otherwise -> digs |
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+ | |||

+ | problem_17 xs = sum . map (length . spell) $ xs |
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</haskell> |
</haskell> |
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## Revision as of 15:43, 1 April 2007

## Contents |

## 1 Problem 11

What is the greatest product of four numbers on the same straight line in the 20 by 20 grid?

Solution:

problem_11 = undefined

## 2 Problem 12

What is the first triangle number to have over five-hundred divisors?

Solution:

problem_12 = head $ filter ((> 500) . nDivisors) triangleNumbers where triangleNumbers = scanl1 (+) [1..] nDivisors n = product $ map ((+1) . length) (group (primeFactors n)) primes = 2 : filter ((== 1) . length . primeFactors) [3,5..] primeFactors n = factor n primes where factor n (p:ps) | p*p > n = [n] | n `mod` p == 0 = p : factor (n `div` p) (p:ps) | otherwise = factor n ps

## 3 Problem 13

Find the first ten digits of the sum of one-hundred 50-digit numbers.

Solution:

nums = ... -- put the numbers in a list problem_13 = take 10 . show . sum $ nums

## 4 Problem 14

Find the longest sequence using a starting number under one million.

Solution:

problem_14 = undefined

## 5 Problem 15

Starting in the top left corner in a 20 by 20 grid, how many routes are there to the bottom right corner?

Solution:

problem_15 = undefined

## 6 Problem 16

What is the sum of the digits of the number 2^{1000}?

Solution:

dsum 0 = 0 dsum n = let ( d, m ) = n `divMod` 10 in m + ( dsum d ) problem_16 = dsum ( 2^1000 )

## 7 Problem 17

How many letters would be needed to write all the numbers in words from 1 to 1000?

Solution:

-- not a very concise or beautiful solution, but food for improvements :) names = concat $ [zip [(0, n) | n <- [0..19]] ["", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight" ,"Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen" ,"Sixteen", "Seventeen", "Eighteen", "Nineteen"] ,zip [(1, n) | n <- [0..9]] ["", "Ten", "Twenty", "Thirty", "Fourty", "Fifty", "Sixty", "Seventy" ,"Eighty", "Ninety"] ,[((2,0), "")] ,[((2, n), look (0,n) ++ " Hundred and") | n <- [1..9]] ,[((3,0), "")] ,[((3, n), look (0,n) ++ " Thousand") | n <- [1..9]]] look n = fromJust . lookup n $ names spell n = unwords $ if last s == "and" then init s else s where s = words . unwords $ map look digs' digs = reverse . zip [0..] . reverse . map digitToInt . show $ n digs' = case lookup 1 digs of Just 1 -> let [ten,one] = filter (\(a,_) -> a<=1) digs in (digs \\ [ten,one]) ++ [(0,(snd ten)*10+(snd one))] otherwise -> digs problem_17 xs = sum . map (length . spell) $ xs

## 8 Problem 18

Find the maximum sum travelling from the top of the triangle to the base.

Solution:

problem_18 = undefined

## 9 Problem 19

How many Sundays fell on the first of the month during the twentieth century?

Solution:

problem_19 = undefined

## 10 Problem 20

Find the sum of digits in 100!

Solution:

problem_20 = let fac n = product [1..n] in foldr ((+) . Data.Char.digitToInt) 0 $ show $ fac 100

Alternate solution, summing digits directly, which is faster than the show, digitToInt route.

dsum 0 = 0 dsum n = let ( d, m ) = n `divMod` 10 in m + ( dsum d ) problem_20' = dsum . product $ [ 1 .. 100 ]