Greek Anthology Book XIV: 6. - Problem
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Problem
- "Best of clocks, how much of the day is past?"
- There remain twice two-thirds of what is gone.
Solution
Let $h$ be the number of hours that have passed.
A day of $12$ hours is assumed.
Then:
\(\ds 12 - h\) | \(=\) | \(\ds 2 \times \dfrac 2 3 h\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 36 - 3 h\) | \(=\) | \(\ds 4 h\) | clearing fractions | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 36\) | \(=\) | \(\ds 7 h\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds h\) | \(=\) | \(\ds \dfrac {36} 7\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 5 \frac 1 7\) |
So $5 \frac 1 7$ hours have passed, and $6 \frac 6 7$ remain.
$\blacksquare$
Sources
- 1918: W.R. Paton: The Greek Anthology Book XIV ... (previous) ... (next): $6$. -- Problem
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Metrodorus and the Greek Anthology: $34$