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Little explorations with HP calculators (no Prime)
03-15-2017, 11:11 PM (This post was last modified: 10-17-2017 03:40 PM by pier4r.)
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Little explorations with HP calculators (no Prime)
Digression
So, after years I have a bit of time to reuse the hp50g and I'm pretty rusty. Not that I was great before, but surely I was a bit more fluent in the library of built in functions of my hp50g.

While I know that already this forum and the old forum on this site have plenty of interesting topics to explore, without checking also comp.sys.hp48, the official support forum by HP and other online forums (if still the links that I have are valid), it is a bit time consuming to find them among other topics. It is a great tasks but not now.

So as a quick but not so trivial (i.e: I can solve them even when I'm tired only with my mind, not even paper) source of math problems to solve with the help of the calculator, I turned to brilliant.org that now, as in 2013, is the only site of its type that I know. (To find similar collections of problems one has to dig forums, like here)
End of digression

From here, every time I cannot solve a problem that I think could be solved with the calculator, I ask for your help if you don't mind. Also, since the new forum let posts grow, I hope to use just this thread to accumulate stuff without creating multiple ones.

So first problem. (neat the inclusion of MathJax, kudos to the admin!)
Quote: \[ \Large \begin{array} {c c c } & & \color{green}{C} \\ + & \color{green}{C} & \color{green}{C} \\
\hline & \color{purple}{D} & 4 \\ \end{array} \]

In this cryptogram, \(\color{green}{C}\) and \(\color{purple}{D}\) represent two different digits. What is the value of \( \color{green}{C} \)?

Now I solved this manually, because I was not able to find a way to let the calculator solve this.

manually I ended up recognizing that C cannot be 2, so it has to be " 2*c mod 10 = 4 " , once found C, the rest is fine.
For the calculator, though, I was using something like:

\[ c + 10c + c = 10d + 4 \]

But that is obviously not enough for the solver. So, how would you solve it?

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Little explorations with HP calculators (no Prime) - pier4r - 03-15-2017 11:11 PM



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