# Pefsu problem explanation

Problem no. 12 from Moscow Mathematical Papyrus:

1. Example of calculation of $$13$$ heqats of grain
2. If someone says to you: Take $$13$$ heqats of grain to make them into $$18$$ jugs of beer
3. Note that the amount of grain for $$1$$ jug is $$2\frac{1}{6}$$.
4. Reckon with $$2\frac{1}{6}$$ in order to find $$13$$.
5. The result is $$6$$ times.
6. Reckon with $$6$$ to find $$18$$.
7. The result is $$3$$ pefsu and this is the solution.

The only ratio that is sufficient for solving these problems is the following:

$$\text{pefsu} = \frac{\text{number of jugs of beer}}{\text{number of heqats of grain}}$$

The problem is essentially looking for the pefsu of those $$18$$ beers that need to be made. I am stuck at the line $$6$$ where I don't understand why the author divides $$18$$ by $$6$$ to get a pefsu of $$3$$. With that logic it would be that not all $$13$$ initial heqats of grain were used to produce $$18$$ jugs of beer since the pefsu formula would then be $$3 = \frac{18}{6}$$, i.e. only $$6$$ heqats were used to produce the needed number of beer. Anyone can explain?

Here is a reference from a book. The book specifies this as a simple problem, so I am looking for an answer, because it makes no sense.