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Little explorations with HP calculators (no Prime)
03-24-2017, 01:45 PM (This post was last modified: 03-24-2017 03:23 PM by pier4r.)
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RE: Little explorations with the HP calculators
Quote:Brilliant.org
We know
\[a^{m+n}=a^m \times a^n\]
But there are some students who claim that
\[a^{m+n}=a^m+a^n\]
which is obviously incorrect. However, it is true for some triplets of integers, \((a,m,n)\). If \(a\in[0,10]\) and \(m,n\in[1,10]\), then find the number of possible triplets for which \(a^{m+n}=a^m+a^n\).

Is there a way to solve this without using a wrapping program? (hp 50g)

I'm trying around some functions (e.g: MSLV) with no luck, so I post this while I dig more on the manual, and on search engines focused on "site:hpmuseum.org" or "comp.sys.hp48" (the official hp forums are too chaotic, so I won't search there, although they store great contributions as well).

edit: I don't mind inserting manually new starting values, I mean that there is a function to find at least one solution, then the user can find the others changing the starting values.

Edit. It seems that the numeric solver for one equation can do it, one has to set values to the variables and then press solve, even if one variable has already a value (it was not so obvious from the manual, I thought that variables with given values could not change it). The point is that one variable will change while the others stay constant. In this way one can find all the solutions.

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RE: Little explorations with the HP calculators - pier4r - 03-24-2017 01:45 PM



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