challenge for programmable calculators

[Han]

(12-26-2013 06:19 AM)Thomas Klemm Wrote:  Or just have a look at the roots: {0.04974, 0.857089, 1.50476, 15.5884}
Since \(f(1)=1\) it's obvious that \(f(n)<0\) for n in {2,...,15}.
To show this for n = 15 with your method isn't possible.
In this case we're lucky that we don't have to do that.

The obviousness of \( f(n) < 0 \) for \( n \in \{ 2, 3, \dotsm, 15 \} \) only comes after having used a solver, a decent graphing program, or the closed formulas for solutions to quartic polynomials. However, the mathematics involved in coming up with those formulas is not trivial. So if you did not use any software (solver or grapher), I fail to see how you can conclude "obviousness."

I'm a bit confused here. I thought the point was (in trying to go for the fastest program) to try to reduce the complexity as much as possible prior to doing any programming. My point was that in the process of reducing the complexity by using mathematics (i.e. no programming or technology), we essentially solved the problem and made the programming challenge unnecessary.

The original problem has been reduced to proving \( a = 1 \), which can be done without any solver and just basic calculus. Cruff's equation 3 is entirely unecessary. All we need (which Cruff showed) are:
\[ abc = 11a+b+1 \quad \text{and} \quad a+b+c = 9 \]
From these two formulas, we can deduce \( 1 \le a,b,c \le 7\) and that
\[ab^2+(1+a^2−9a)b+11a+1=0, \]
a point that you brought up in a previous post. In solving for \( b \) in \(ab^2+(1+a^2−9a)b+11a+1=0 \), \( b \) being a positive integer forces the condition that the descriminant must at least be non-negative. This leads to the conclusion that \( a=1 \) is the only possibility (through basic calculus), and the solution to the original programming challenge follows easily.

We never need to consider values of \( a > 7 \) since it can be proved that \( 1 \le a \le 7 \). I am also not seeing how luck factors in, either, given that everything was deduced mathematically.

Edit: Given that \( a \) can only be 7 possible integer values, brute-force-by-hand would also lead to the conclusion that the descriminant is positive only for \( a = 1 \). However Don Shepherd had interest in a non-brute force solution.