challenge for programmable calculators

[Gerson W. Barbosa]

(12-24-2013 05:43 PM)cruff Wrote:  Substitution of the seven possible values of \(a\) into equation 3 and using the unique factorization theorem for the resulting values of the numerator \(11a+1\) limit the search space for possible values of \(n\) that result in integers for \(b\). Back substitution into equation 1 produces the two solutions and a number of contradictions without too much effort. I wish I could have proved that \(a\) must be \(1\), which would have further limited the search for n by a factor of 7, but that escapes my aging brain at the moment.

Using two of your conclusions,
\begin{align}
c &= \frac{11+\frac{b+1}a}b\\\end{align}
and
\begin{align}
a+b+c=9\end{align}
one can write the following RPL program,

Code:

%%HP: T(3)A(D)F(.);
\<< 1. 7.
  FOR a 1. 7.
    FOR b b 1. + a / 11. + b / DUP FP NOT NOT { DROP } { DUP 9. a - b - == { a 100. * b 10. * + + } { DROP } IFTE } IFTE
    NEXT
  NEXT
\>>

which finds both solutions in 0.6675 seconds on the real hp 50g. That's about a 22-time improvement over the plain brute-force solution. This would have been even faster if all your conclusions were taken into account (if a program would be necessary at all).
Quite impressive!