Greek Anthology Book XIV: 51. - Problem

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Problem

$A$. I have what the second has and the third of what the third has.
$B$. I have what the third has and the third of what the first has.
$C$. And I have ten minae and the third of what the second has.


Solution

Let $a$, $b$ and $c$ (respectively) be what $A$, $B$ and $C$ have in minae.

We have:

\(\ds a\) \(=\) \(\ds b + \frac c 3\)
\(\ds b\) \(=\) \(\ds c + \frac a 3\)
\(\ds c\) \(=\) \(\ds 10 + \frac b 3\)
\(\ds \leadsto \ \ \) \(\ds 3 a\) \(=\) \(\ds 3 b + c\) multiplying through by $3$
\(\ds 3 b\) \(=\) \(\ds 3 c + a\)
\(\ds 3 c\) \(=\) \(\ds 30 + b\)
\(\ds \leadsto \ \ \) \(\ds 3 b\) \(=\) \(\ds 30 + b + a\)
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds 2 b - 30\)
\(\ds \leadsto \ \ \) \(\ds 3 \paren {2 b - 30}\) \(=\) \(\ds 3 b + 10 + \frac b 3\)
\(\ds \leadsto \ \ \) \(\ds 18 b - 270\) \(=\) \(\ds 9 b + 30 + b\)
\(\ds \leadsto \ \ \) \(\ds 8 b\) \(=\) \(\ds 300\)
\(\ds \leadsto \ \ \) \(\ds b\) \(=\) \(\ds \dfrac {300} 8\)
\(\ds \) \(=\) \(\ds 37 \frac 1 2\)
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds 2 \paren {37 \frac 1 2} - 30\)
\(\ds \) \(=\) \(\ds 75 - 30\)
\(\ds \) \(=\) \(\ds 45\)
\(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds 10 + \frac 1 3 \times {37 \frac 1 2}\)
\(\ds \) \(=\) \(\ds 10 + \frac 1 2 \frac {75} 3\)
\(\ds \) \(=\) \(\ds 10 + \frac {25} 2\)
\(\ds \) \(=\) \(\ds 22 \frac 1 2\)

So:

$A$ has $45$ minae
$B$ has $37 \frac 1 2$ minae
$C$ has $22 \frac 1 2$ minae.

$\blacksquare$


Sources

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