Greek Anthology Book XIV: 12. - Problem

From ProofWiki
Jump to navigation Jump to search

Problem

Croesus the king dedicated six bowls weighing six minae,
each one drachm heavier than the other.


Solution

It is known that there are $100$ drachms to the mina.


The weights of the bowls form an arithmetic sequence.

Let $a$ denote the weight of the lightest bowl in drachms.

Let $n$ denote the number of bowls.

Let $d$ denote the common difference between the weights in drachms of consecutive bowls arranged in order of weight.

Let $t$ denote the total weight of all bowls in drachms.


We are given:

\(\ds n\) \(=\) \(\ds 6\)
\(\ds t\) \(=\) \(\ds 6 \times 100 = 600\)
\(\ds d\) \(=\) \(\ds 1\)


So:

\(\ds t\) \(=\) \(\ds n \paren {a + \frac {n - 1} 2 d}\) Sum of Arithmetic Sequence
\(\ds \leadsto \ \ \) \(\ds 600\) \(=\) \(\ds 6 \paren {a + \frac {6 - 1} 2}\)
\(\ds \) \(=\) \(\ds 6 a + 15\) simplifying
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds \dfrac {600 - 15} 6\)
\(\ds \) \(=\) \(\ds \dfrac {585} 6\)
\(\ds \) \(=\) \(\ds 97 \frac 1 2\)


So the lightest bowl weighs $97 \frac 1 2$ drachms, and it follows that the weights of each bowl in drachms is:

$97 \frac 1 2, 98 \frac 1 2, 99 \frac 1 2, 100 \frac 1 2, 101 \frac 1 2, 102 \frac 1 2$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Greek_Anthology_Book_XIV:_12._-_Problem&oldid=480682"