How can I determine the base number system $x$ of two decimal numbers using an equation?
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3It would be easier to find the answer to $59_{x}=89_{10}$ and then check if that worked for your expression – Henry Aug 19 '22 at 23:07
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@Henry: You can take the integer part if you know up front that $x$ is an integer. However, there also turns out to be a fractional-base solution, as noted in my answer. – Dan Aug 19 '22 at 23:29
2 Answers
Recall that a place-value numeral is just a polynomial (or a Laurent polynomial if there are digits after the point) at the base. Thus, "59.5" in base $x$ means $5x + 9 + 5x^{-1}$. So you just need to solve:
$$5x + 9 + 5x^{-1} = 89.3125$$ $$5x^2 + 9x + 5 = 89.3125x$$ $$5x^2 - 80.3125x + 5 = 0$$
Or, multiplying by $\frac{16}{5}$ so that everything is an integer
$$16x^2 - 257x + 16 = 0$$ $$x = \frac{257 \pm \sqrt{(-257)^2 - 4(16)(16)}}{2(16)}$$ $$x = \frac{257 \pm \sqrt{65025}}{32}$$ $$x = \frac{257 \pm 255}{32}$$ $$x \in \lbrace \frac{1}{16}, 16 \rbrace$$
Although fractional bases are a thing, they're way less common than integer bases, so 16 is probably the intended answer.
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2It's interesting that the fractional base is the reciprical of the digital base. Then I realized that is precisely because the number $59.5$ is palindromic (So $5\times 16 + 9 +\frac 5{16} = 5\times \frac 1{16} + 9 + \frac 5{\frac 1{16}}$). Be interesting to find the fraction base of a non-palindromic number. Be interesting to figure out what integers are and arenot expressible by a fractional base that is not of the form $\frac 1b$. – fleablood Aug 20 '22 at 02:51
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@fleablood: It may depend on whether there are restrictions on the digits. For example $31.2_{2/3} = 6_{10}$ but you also have $90_{2/3} = 6_{10}$ and $0.4_{2/3} = 6_{10}$ and of course $6_{2/3} = 6_{10}$ – Henry Aug 20 '22 at 13:07
Well, a freebie... as $5 < 8$ we know that $x > 10$.
But basically we have $5x + 9 + \frac 5x = 89.3125$
Since $\frac 5x < 1$ we have the $5x +9$ must be an integer and $5x+9 = 89$ and so $x =16$.
That's all. That's the only thing it could be.
But that's not enough. We have to verify that $x=16$ is possible. After all, just because we asked a question doesn't mean there is actually an answer.
So we need to verify that $\frac 5{16} = 0.3125$ which ... it does.
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