Problem 9 [Language]

[jamiephillips0000]

Hi Guys
Really had fun with Problem 9

def divider(nr : Int) ={ 2000 - (2_nr)}
def numerator(nr :Int) = {1000_1000 - (2000 * nr)}
def isPos(nr : Int) = {1000*1000 - (2000 * nr) > 0}
val res = (1 to 1000).filter(x => x< 1000 && isPos(x) && numerator(x) % divider(x)==0)

println (res(0) + " --" + res(1)  + " -- " + math.sqrt(res(0)*res(0) + res(1)*res(1)).toInt)

res(0) * res(1) * math.sqrt(res(0)*res(0) + res(1)*res(1)).toInt

One suggestion though - It would be nice to know the format in which the answer should be given e.g. i had to infer from the answer of @LucDupAtGitHub that the product of all three was required. i had the correct solution but was not sure of the format of the answer. This is not a criticism just a remark ;-)

[eloots]

Hi Jamie,

Indeed a fun problem to tackle !

As for the problem statement; indeed: it seems I forgot to add the last sentence to the problem description which is: Find the product abc

Sorry for that ;-) Of course, in case of doubt, there’s always the source at https://projecteuler.net/problems https://projecteuler.net/problems.

Regards, Eric

On 12-dec.-2014, at 09:42, Jamie Phillips notifications@github.com wrote:

Hi Guys
Really had fun with Problem 9

def divider(nr : Int) ={ 2000 - (2nr)}
def numerator(nr :Int) = {10001000 - (2000 * nr)}
def isPos(nr : Int) = {1000*1000 - (2000 * nr) > 0}
val res = (1 to 1000).filter(x => x< 1000 && isPos(x) && numerator(x) % divider(x)==0)

println (res(0) + " --" + res(1) + " -- " + math.sqrt(res(0)_res(0) + res(1)_res(1)).toInt)

res(0) * res(1) * math.sqrt(res(0)_res(0) + res(1)_res(1)).toInt
One suggestion though - It would be nice to know the format in which the answer should be given e.g. i had to infer from the answer of @LucDupAtGitHub https://github.com/LucDupAtGitHub that the product of all three was required. i had the correct solution but was not sure of the format of the answer. This is not a criticism just a remark ;-)

—
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[eloots]

Jamie,

By the way; a related problem to this one is:

https://projecteuler.net/problem=39 <https://projecteuler.net/problem=39>

It’s not in the set of problems include in the repo.

You may want to have a crack at it. The core stuff you need is already there.

If you’d like to have it in included in the repo, let me know and I’ll add it.

Cheers, Eric

On 12-dec.-2014, at 09:42, Jamie Phillips notifications@github.com wrote:

Hi Guys
Really had fun with Problem 9

def divider(nr : Int) ={ 2000 - (2nr)}
def numerator(nr :Int) = {10001000 - (2000 * nr)}
def isPos(nr : Int) = {1000*1000 - (2000 * nr) > 0}
val res = (1 to 1000).filter(x => x< 1000 && isPos(x) && numerator(x) % divider(x)==0)

println (res(0) + " --" + res(1) + " -- " + math.sqrt(res(0)_res(0) + res(1)_res(1)).toInt)

res(0) * res(1) * math.sqrt(res(0)_res(0) + res(1)_res(1)).toInt
One suggestion though - It would be nice to know the format in which the answer should be given e.g. i had to infer from the answer of @LucDupAtGitHub https://github.com/LucDupAtGitHub that the product of all three was required. i had the correct solution but was not sure of the format of the answer. This is not a criticism just a remark ;-)

—
Reply to this email directly or view it on GitHub #17.

[mverbist]

And this is just using brute-force, without Erics tip:

      (for {
          a <- 1 to 999
          b <- a to 999
          c <- b to 999 filter (c1 => a + b + c1 == 1000 && a*a + b*b == c1*c1)
      } yield (a * b * c)).head

A triple for-comprehension with the condition as a filter on the third.
We yield the product, and then take the first we find.

[ghost]

very nice, Michel!

On Fri, Jan 9, 2015 at 11:00 AM, Verbist Michel notifications@github.com
wrote:

And this is just using brute-force, without Erics tip:

  (for {
      a <- 1 to 999
      b <- a to 999
      c <- b to 999 filter (c1 => a + b + c1 == 1000 && a*a + b*b == c1*c1)
  } yield (a * b * c)).head

A triple for-comprehension with the condition as a filter on the third.
We yield the product, and then take the first we find.

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#17 (comment)
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