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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.

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