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I want to count total number of the natural solutions (different from 0) of the equation $2x + 3y + z = 100$, but don't know how. How can I calculate it using Mathematica? I tried:

Solve[{2*x + 3*y + z == 100, x > 0, y > 0, z > 0}, {x, y, z}, Integers]
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This should work:

sols = Solve[{2*x + 3*y + z == 100, x > 0, y > 0, z > 0}, 
        {x, y, z}, Integers];
Length@sols
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There is an especially useful function for this kind of task: FrobeniusSolve[{a, b, c}, d] for finding the list of all solutions to the equation a x + b y + c z == d, where a,b,c are given positive integers and d is an integer, while x,y,z are non-negative integers to be found. There are many solutions (884 of them):

FrobeniusSolve[{2, 3, 1}, 100] // Short

{{0, 0, 100}, {0, 1, 97}, {0, 2, 94}, {0, 3, 91}, <<876>>, {48, 0, 4},
 {48, 1, 1},  {49, 0, 2}, {50, 0, 0}}

Note that I used Short which omitted 876 solutions. If you need only positive solutions you can use e.g. DeleteCases to get lists free of 0:

DeleteCases[ FrobeniusSolve[{2, 3, 1}, 100], {___, 0, ___}] // Length

 784

Edit

FrobeniusSolve is not a superfluous function since it is much more efficient than an adequate use of Solve. Consider e.g. the analogous equation 2 x + 3 y + z == 1000. We compare the timings of solving the same equation:

Solve[{2*x + 3*y + z == 1000, x >= 0, y >= 0, z >= 0}, {x, y, z}, Integers] //
Length // AbsoluteTiming
FrobeniusSolve[ {2, 3, 1}, 1000] // Length // AbsoluteTiming

 {76.9510000, 83834}
  {0.1900000, 83834}

This huge efficiency difference doesn't really change if we are to find positive solutions:

Solve[{2*x + 3*y + z == 1000, x > 0, y > 0, z > 0}, {x, y, z}, Integers] //
Length // AbsoluteTiming
DeleteCases[ FrobeniusSolve[{2, 3, 1}, 1000], {___, 0, ___}] // Length // AbsoluteTiming

 {77.9720000, 82834}
  {0.2420000, 82834}

Moreover, the lists of solutions have the same orderings; e.g.:

DeleteCases[ FrobeniusSolve[{2, 3, 1}, 200], {___, 0, ___}] === 
Solve[{2*x + 3*y + z == 200, x > 0, y > 0, z > 0}, {x, y, z}, Integers][[All, All, 2]]

True
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  • 2
    $\begingroup$ +1 for my favorite Frobenius* functions! $\endgroup$ – Silvia Sep 15 '12 at 18:48
  • $\begingroup$ @Silvia Thanks ! Functions like FrobeniusSolve are indeed really useful. I think its documentation might be a bit more comprehensive. $\endgroup$ – Artes Sep 15 '12 at 19:56
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There are also power series methods for counting these.

SeriesCoefficient[
 x^(1 + 2 + 3)/(1 - x^1)*1/(1 - x^2)*1/(1 - x^3), {x, 0, 100}]

(* Out[118]= 784 *)

See also "Supplement to 'Perplexities Related to Fourier's 17 Line Problem'."

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