challenge for programmable calculators

[cruff]

(12-21-2013 06:15 PM)Don Shepherd Wrote:  A non-brute-force solution would greatly enlighten this rainy day.

I believe I have the proof using a bit of basic number theory. It has been 30 years or so since I took my number theory class, so if anyone spots an error, please provide a correction! By the way, does anyone know of a way to number the equations nicely near the margin as is done in texts?

Definition 1: Let \(a, b, c\) be integers such that: \(1 \le a,b,c \le 9\)
Equation 1:
\[abc(a+b+c) = 100a + 10b + c\]
Note that \(a+b+c\) is present on both sides and rewrite the equation a bit:
\begin{align}
abc(a+b+c) &= 99a+9b+(a+b+c)\\
abc &= \frac{9(11a+b)+(a+b+c)}{(a+b+c)}\\
abc &= \frac{9(11a+b)}{(a+b+c)} + 1\end{align}
Equation 2:
\[abc-1 = \frac{9(11a+b)}{a+b+c}\]
Since we are dealing exclusively with positive non-zero integers, both \(abc-1\) and \(\frac{9(11a+b)}{a+b+c}\) must both be integral. This means that the term \(a+b+c\) must be a factor of \(9(11a+b)\). Because \(9(11a+b)\) is > 1 by definition 1, the unique factorization theorem states that it is a product of primes and the factorization is unique. Thus the factors are \(3, 3^2\) and \(11a + b\).

We won't tackle the determination of if \(11a + b\) is prime, but will instead show a contradiction results:
\begin{align}
a+b+c &= 11a+b\\
c &= 10a\end{align}Which contradicts definition 1.

Next we examine if \(a+b+c=3\). This implies that \(a=1\), \(b=1\) and \(c=1\), which leads to a contradiction of equation 1.

This leaves the \(a+b+c=9\) possibility. Substituting into equation 2 produces:
\begin{align}
abc-1 &= \frac{9(11a+b)}{(a+b+c)}\\
abc-1 &= \frac{9(11a+b)}{9}\\
abc-1 &= 11a+b\end{align}
Now we solve for \(c\):
\begin{align}
abc &= 11a+b+1\\
c &= \frac{11a+b+1}{ab}\\
c &= \frac{11a}{ab}+\frac{b}{ab}+\frac1{ab}\\
c &= \frac{11}b+\frac{b+1}{ab}\\
c &= \frac{11}b+\frac{\frac{b+1}a}b\\
c &= \frac{11+\frac{b+1}a}b\\\end{align}
Since we are dealing only with non-zero positive integers, this means b must evenly divide \(11+\frac{b+1}a\). Therefore, there exists some integer \(n\), \(n > 0\), such that:
\[11+\frac{b+1}a = nb\]
Solve for \(b\):
\begin{align}
11+\frac{b+1}a &= nb\\
11a+b+1 &= anb\\
11a+1 &= (an-1)b\\
\frac{11a+1}{an-1} &= b\end{align}
If \(a+b+c=9\), this implies that \(1 \le a,b,c \le 7\) and thus:
Equation 3:
\[1 \le \frac{11a+1}{an-1} \le 7\]
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.