Five pirates found a treasure chest full of gold coins and divided them among themselves. It is known that:

- What the first pirate received is equivalent to half of what the other four received in total.
- What the second pirate received is equivalent to a third of what the other four received in total.
- What the third pirate received is equivalent to a quarter of what the other four received in total.
- What the fourth pirate received is equivalent to a fifth of what the other four received in total.

If the fifth pirate received 90 coins, say how many coins were in the chest before the division.
Solution

Let $a, b, c, d$, $e$ be the number of coins received by the five pirates. The total number of coins is $S=a+b+c+d+e$. The first pirate received half of what the other four received in total, that is, $a=(b+c+d+e) / 2=(S-a) / 2$. Therefore, $a=S / 3$. The second pirate received a third of what the other four received in total, that is, $b=(a+c+d+e) / 3=(S-b) / 3$. Therefore, $b=S / 4$. The third pirate received a quarter of what the other four received in total, that is, $c=(a+b+d+e) / 4=(S-c) / 4$. Therefore, $c=S / 5$. The fourth pirate received a fifth of what the other four received in total, that is, $d=(a+b+c+e) / 5=(S-d) / 5$. Therefore, $d=S / 6$. Thus, the last pirate received

$$
90=e=S-a-b-c-d=S-\frac{S}{3}-\frac{S}{4}-\frac{S}{5}-\frac{S}{6}=\frac{S}{20}
$$

Therefore, the chest had $S=1800$ coins before the division.