Greek Anthology Book XIV: 12. - Problem

Problem

Croesus the king dedicated six bowls weighing six minae,

each one drachm heavier than the other.

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:

$\ds t$

$=$

$\ds n \paren {a + \frac {n - 1} 2 d}$

$\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:

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

$\blacksquare$

1918: W.R. Paton: The Greek Anthology Book XIV ... (previous) ... (next): $12$. -- Problem