# Euler problems/181 to 190

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< Euler problems(Difference between revisions)

Robinrobin (Talk | contribs) ((183) If we must have a solution here, let's at least have a decent one.) |
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<haskell> |
<haskell> |
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− | pmax x a=a*(log x-log a) |
+ | -- Does the decimal expansion of p/q terminate? |

− | tofloat x=encodeFloat x 0 |
+ | terminating p q = 1 == reduce [2,5] (q `div` gcd p q) |

− | fun x= |
+ | where reduce [] n = n |

− | div n1 $gcd n1 x |
+ | reduce (x:xs) n | n `mod` x == 0 = reduce (x:xs) (n `div` x) |

− | where |
+ | | otherwise = reduce xs n |

− | e=exp 1 |
+ | |

− | n=floor(fromInteger x/e) |
+ | -- The expression (round $ fromIntegral n / e) computes the integer k |

− | n1=snd.maximum$[(b,a)|a<-[n..n+1],let b=pmax (tofloat x) (tofloat a)] |
+ | -- for which (n/k)^k is at a maximum. |

− | n `splitWith` p = doSplitWith 0 n |
+ | answer = sum [if terminating n (round $ fromIntegral n / e) then -n else n |

− | where doSplitWith s t |
+ | | n <- [5 .. 10^4]] |

− | | p `divides` t = doSplitWith (s+1) (t `div` p) |
+ | where e = exp 1 |

− | | otherwise = (s, t) |
+ | |

− | d `divides` n = n `mod` d == 0 |
+ | main = print answer |

− | funD x |
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− | |is25 k=(-x) |
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− | |otherwise =x |
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− | where |
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− | k=fun x |
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− | is25 x |
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− | |s==1=True |
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− | |otherwise=False |
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− | where |
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− | s=snd(splitWith (snd (splitWith x 2)) 5) |
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− | problem_183 =sum[funD a|a<- [5..10000]] |
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</haskell> |
</haskell> |

## Revision as of 20:27, 24 February 2008

## 1 Problem 181

Investigating in how many ways objects of two different colours can be grouped.

Solution: This was my code, published here without my permission nor any attribution, shame on whoever put it here. Daniel.is.fischer

## 2 Problem 182

RSA encryption.

Solution:

fun a1 b1 = sum [ e | e <- [2..a*b-1], gcd e (a*b) == 1, gcd (e-1) a == 2, gcd (e-1) b == 2 ] where a=a1-1 b=b1-1 problem_182=fun 1009 3643

## 3 Problem 183

Maximum product of parts.

Solution:

-- Does the decimal expansion of p/q terminate? terminating p q = 1 == reduce [2,5] (q `div` gcd p q) where reduce [] n = n reduce (x:xs) n | n `mod` x == 0 = reduce (x:xs) (n `div` x) | otherwise = reduce xs n -- The expression (round $ fromIntegral n / e) computes the integer k -- for which (n/k)^k is at a maximum. answer = sum [if terminating n (round $ fromIntegral n / e) then -n else n | n <- [5 .. 10^4]] where e = exp 1 main = print answer