Greek Anthology Book XIV: 13. - Problem
Jump to navigation
Jump to search
Problem
- We both of us together weigh twenty minae, I, Zethus, and my brother;
- and if you take the third part of me and the fourth part of Amphion here, you will find it makes six,
- and you will have found the weight of our mother.
Solution
Let $z$ be the weight in minae of Zethus.
Let $a$ be the weight in minae of Amphion.
We have:
| \(\text {(1)}: \quad\) | \(\ds z + a\) | \(=\) | \(\ds 20\) | |||||||||||
| \(\text {(2)}: \quad\) | \(\ds \frac z 3 + \frac a 4\) | \(=\) | \(\ds 6\) | |||||||||||
| \(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds 4 z + 3 a\) | \(=\) | \(\ds 72\) | multiplying $(2)$ by $12$ to clear fractions | |||||||||
| \(\text {(4)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds 4 z + 4 a\) | \(=\) | \(\ds 80\) | multiplying $(1)$ by $4$ | |||||||||
| \(\text {(5)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds a\) | \(=\) | \(\ds 8\) | subtracting $(3)$ from $(4)$ | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds 12\) | substituting for $a$ in $(1)$ and simplifying |
So:
- Zethus weighs $12$ minae
- Amphion weighs $8$ minae
and (presumably):
- their mother weighs $6$ minae.
$\blacksquare$
Sources
- 1918: W.R. Paton: The Greek Anthology Book XIV ... (previous) ... (next): $13$. -- Problem