How To Solve An Equation That Has "Int" In The MixPrev TopicNext Topic

• mitachoo

  

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  Jul 4, 2019, 1:53 pm

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  I have an odd math question, hopefully someone can point me in the right direction.

  I have an equation that looks like this:

  a = (bx + c) - int{ (bx + c)/n } * n

  How the (expletive) do I solve for x, since x has a mix of Int and non-Int?  Is there some way to "convert" the Int portion of the equation to non-Int?
• cottoneyedjoe

  

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  Jul 4, 2019, 4:13 pm

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  Assuming n is an integer and k is either an integer or non-integer, the expression int{ k/n }*n is an integer bounded between a minimum of int{k} - n + 1 and a maximum of int{k}. That is a total of n different values the expression can take on, so to do it algebraically, you could replace k with bx+c and solve n different equations.

  What you will find is that for any given value of "a" that is at least 0 and strictly less than n, there are infinitely many values of x that make the equation true. If "a" is less than 0 or at least n, the equation will have no solutions.

  Probably not the answer you were hoping for!
• mitachoo

  

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  Jul 4, 2019, 7:46 pm

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  In response to cottoneyedjoe

  Quote: Originally posted by cottoneyedjoe on Jul 4, 2019

  Assuming n is an integer and k is either an integer or non-integer, the expression int{ k/n }*n is an integer bounded between a minimum of int{k} - n + 1 and a maximum of int{k}. That is a total of n different values the expression can take on, so to do it algebraically, you could replace k with bx+c and solve n different equations.

  What you will find is that for any given value of "a" that is at least 0 and strictly less than n, there are infinitely many values of x that make the equation true. If "a" is less than 0 or at least n, the equation will have no solutions.

  Probably not the answer you were hoping for!

  Thanks cottoneyedjoe, in a roundabout way I figured there was an infinte number of solutions in that I had resorted to creating a Macro in Excel to iterate on solutions but only up to a certain value of a.  I hoped there might be a better way, but if infinity is the answer, there's naught I can do except bound it as I've done already.

   

  Thanks again!
• JADELottery

  

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  Jul 4, 2019, 10:04 pm

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  Solveable, yes.

  If ¦a¦ < ¦n¦

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• lakerben

  

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  Jul 5, 2019, 12:08 am

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  In response to JADELottery

  Quote: Originally posted by JADELottery on Jul 4, 2019

  Solveable, yes.

  If ¦a¦ < ¦n¦

  Happy  4th jade!  Good to see you on the site!

  I'm sitting here listening to Merle Haggard sipping some coffee!

   

  
• JADELottery

  

  Native American Eagle Sun

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  Jul 5, 2019, 12:54 am

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  In response to lakerben

  Quote: Originally posted by lakerben on Jul 5, 2019

  Happy  4th jade!  Good to see you on the site!

  Thanks.

  Hope your 4th was great, too.

  Been lurking around mostly due to business travel and vacation.

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• JADELottery

  

  Native American Eagle Sun

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  Jul 5, 2019, 1:10 pm

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  Decimal η Integer Part α & Fractional Part β

  if η = α.β, then int(η) = α.0 and frac(η) = 0.β

  
  Relational Equations

  η = int(η) + frac(η)
  int(η) = η - frac(η)
  frac(η) = η - int(η)

  
  Solution

  a = (b · x + c) - int((b · x + c) / n) · n

  a / n = ((b · x + c) / n) - int((b · x + c) / n)

  a / n = frac((b · x + c) / n)

  if and only if |a| < |n| and 0 ≤ (a / n) < 1, then |ℤ| + (a / n) = (b · x + c) / n

  n · |ℤ| + a = b · x + c

  n · |ℤ| + a - c = b · x

  (n · |ℤ| + a - c) / b = x

  x = (n · |ℤ| + a - c) / b

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