# Difference between revisions of "Euler problems/181 to 190"

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Robinrobin (talk | contribs) ((183) If we must have a solution here, let's at least have a decent one.) |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | pmax x a=a*(log x-log a) |
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+ | -- Does the decimal expansion of p/q terminate? |
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− | tofloat x=encodeFloat x 0 |
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+ | terminating p q = 1 == reduce [2,5] (q `div` gcd p q) |
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− | fun x= |
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+ | where reduce [] n = n |
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− | div n1 $gcd n1 x |
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+ | reduce (x:xs) n | n `mod` x == 0 = reduce (x:xs) (n `div` x) |
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− | where |
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+ | | otherwise = reduce xs n |
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− | e=exp 1 |
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+ | |||

− | n=floor(fromInteger x/e) |
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+ | -- The expression (round $ fromIntegral n / e) computes the integer k |
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− | n1=snd.maximum$[(b,a)|a<-[n..n+1],let b=pmax (tofloat x) (tofloat a)] |
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+ | -- for which (n/k)^k is at a maximum. |
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− | n `splitWith` p = doSplitWith 0 n |
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+ | answer = sum [if terminating n (round $ fromIntegral n / e) then -n else n |
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− | where doSplitWith s t |
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+ | | n <- [5 .. 10^4]] |
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− | | p `divides` t = doSplitWith (s+1) (t `div` p) |
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+ | where e = exp 1 |
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− | | otherwise = (s, t) |
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+ | |||

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

## 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

## 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
```

## 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
```