Greek Anthology Book XIV: 50. - Problem
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Problem
- Throw me in, silversmith, besides the bowl itself,
- the third of its weight,
- and the fourth,
- and the twelfth;
- and casting them into the furnace stir them, and mixing them all up take out, please, the mass, and let it weigh one mina.
Solution
Let $w$ be the weight in minae of the bowl.
We have:
\(\ds w + \frac w 3 + \frac w 4 + \frac w {12}\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 12 w + 4 w + 3 w + w\) | \(=\) | \(\ds 12\) | multiplying through by $12$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 20 w\) | \(=\) | \(\ds 12\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds w\) | \(=\) | \(\ds \dfrac {12} {20}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 3 5\) |
Thus the bowl weighs $\dfrac 3 5$ of a mina.
Assuming $100$ drachmae to the mina, that is, $60$ drachmae.
$\blacksquare$
Sources
- 1918: W.R. Paton: The Greek Anthology Book XIV ... (previous) ... (next): $50$. -- Problem