Problem 7.2. Three pirates were dividing a treasure. The first one got a third of the initial number of coins and one more coin, the second one got a quarter of the initial number of coins and five more coins, the third one got a fifth of the initial number of coins and twenty more coins (thus all the coins were distributed). How many coins were in the treasure?
Answer: 120.

Solution. Let $m$ be the initial total number of coins in the treasure. The pirates collectively took $\left(\frac{m}{3}+1\right)+\left(\frac{m}{4}+5\right)+\left(\frac{m}{5}+20\right)$ coins. Since they divided the entire treasure, this sum equals $m$. Expanding the brackets in the equation

$$
\left(\frac{m}{3}+1\right)+\left(\frac{m}{4}+5\right)+\left(\frac{m}{5}+20\right)=m
$$

and bringing the fractions to a common denominator, we get $\frac{47}{60} m+26=m$, from which $\frac{13}{60} m=26$, that is, $m=120$.