generating least positive integer (mod function) help

[apomb]

Sorry, ya TheGuardian is right...

Thank you TheGuardian for clearing that up for me.

[XXX111]

Thanks apomb and guardian for your excellent help

[Brightguy]

XXX111 @ Thu Apr 08, 2010 4:44 pm wrote:

I almost have the code completed except for one small problem. The program now generates all the possible integers, but is there any way to stop after the first one (also the lowest) and have it display?

Well, actually, you shouldn't have to stop. Since 7,8,9 are coprime (they share no prime divisors) the solution between 1 and 504 will be unique. Remember you want an x that satisfies each constraint, not just one.

[Zren]

Exit works for FOR loops just as good as normal loops.

Turing:

var n : int := 1 loop     exit when %Insert conditions here%     exit when n > maxint     n += 1 end loop put n for i : 1 .. maxint     if %Insert conditions here% then         put i         exit     end if end for

Oh and 504 isn't a proper exit strategy since we're not talking about x rem y = 0 (aka a factor). Your just lucky that the answer is slightly below this though. Then again there might be a theory that says every combination of below that sum will have each combination of remainders...

@Brightguy: is that what your Chinese remainder theorem was pointing at?

[Brightguy]

Zren @ Thu Apr 08, 2010 8:24 pm wrote:

Then again there might be a theory that says every combination of below that sum will have each combination of remainders...

I'm having difficulty parsing this, but yes, the Chinese remainder theorem is the essential theory. It saves you the trouble of checking every possible case by giving you a formula for the answer:

4*8*9*(1/(8*9) mod 7)+5*7*9*(1/(7*9) mod 8)+6*7*8*(1/(7*8) mod 9) mod 7*8*9

However, this requires computing modular inverses, which may not be part of the high school curiculum.