Greek Anthology Book XIV: Metrodorus: 145

From ProofWiki
Jump to navigation Jump to search

Arithmetical Epigram of Metrodorus

$A$. Give me ten minas and I become three times as much as you.
$B$. And if I got the same from you I am five times as much as you.


Solution

Let $a$ minas be the quantity of $A$.

Let $b$ minas be the quantity of $B$.


We have:

\(\text {(1)}: \quad\) \(\ds a + 10\) \(=\) \(\ds 3 \paren {b - 10}\)
\(\ds \leadsto \ \ \) \(\ds a + 10\) \(=\) \(\ds 3 b - 30\)
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds 3 b - 40\)
\(\text {(2)}: \quad\) \(\ds b + 10\) \(=\) \(\ds 5 \paren {a - 10}\)
\(\ds \leadsto \ \ \) \(\ds b + 10\) \(=\) \(\ds 5 a - 50\)
\(\ds \leadsto \ \ \) \(\ds b\) \(=\) \(\ds 5 a - 60\)
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds 3 \paren {5 a - 60} - 40\)
\(\ds \) \(=\) \(\ds 15 a - 180 - 40\)
\(\ds \leadsto \ \ \) \(\ds 14 a\) \(=\) \(\ds 220\)
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds \frac {110} 7\)
\(\ds \) \(=\) \(\ds 15 \frac 5 7\)
\(\ds \leadsto \ \ \) \(\ds b\) \(=\) \(\ds 5 \times \frac {110} 7 - 60\)
\(\ds \) \(=\) \(\ds \frac {550 - 420} 7\)
\(\ds \) \(=\) \(\ds \frac {130} 7\)
\(\ds \) \(=\) \(\ds 18 \frac 4 7\)

So:

$A$ is $15 \dfrac 5 7$
$B$ is $18 \dfrac 4 7$.

$\blacksquare$


Source of Name

This entry was named for Metrodorus.


Sources