## Fractions are Hard!

### 4.2 Shepherd and Sheep Problems

Egyptian fractions also make a wonderful appearance in a set of classic teasers.

An elderly shepherd, with a poor understanding of mathematics, died and left his entire flock of 17 sheep to his three sons. To his first son, whom he admired dearly, he bequeathed half his flock of sheep. To his second son, with whom he was in some favor, one third his flock and to his third son, whom he considered a rogue, one-ninth of his flock. The sons realized that the proportions are strange and that the number 17 is resistant to convenient divisibility, yet they found a clever way to successfully honor their father’s wishes. What was their solution? |

Meanwhile, three daughters of a recently deceased shepherdess faced a similar mathematical predicament. Their mother, very wealthy but also possessing a vague understanding of fractions, had bequeathed her flock of 624 llamas according to the proportions 1/5 to her first daughter, 1/30 to her second daughter, and 1/2670 to her third!
The three daughters managed to fulfill her mother’s wishes. How? |

These classic herd-sharing puzzles have a trick solution (that does not require converting the flock to mutton!)

Think about these problems before reading on.

***

Here are the answers:

The three sons borrowed one sheep from a neighbor to boost the count of their flock to 18. From this flock, the first son took 9 sheep (half), the second 6 (a third), and the third 2 (a ninth). This left one sheep over, which they returned to the neighbor!

The three daughters realized, on the other hand, they needed to borrow 2046 llamas from another source. This boosted the flock to a count of 2670 from which the first daughter took 534 llamas (one fifth), the second 89 llamas (one 30^{th}) and the third 1 llama (one 2670^{th} of the flock). This left 2046 llamas, which they promptly returned.

**Comment: **There is a legal problem with this approach: An estate cannot accrue further debt before distributing its goods. We’ll just set this issue to the side.

Notice that

\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{9}=\dfrac{17}{18}\).

If we rewrite this as \(18\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{9}\right)=17\) we see why boosting to a count of 18 sheep solves the first problem. (Notice too that 18 is lowest common multiple of 2, 3, and 9.)

Similarly \(\dfrac{1}{5}+\dfrac{1}{30}+\dfrac{1}{2670}=\dfrac{624}{2670}\) yields a match with the three-daughter llama puzzle.

We see now that it is easy to create herd-sharing problems of this type: Simply list a set of fractions and represent their sum (hopefully less than 1) as a fraction with denominator the lowest common multiple of the original denominators. For example, the sum

\(\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{2}{15}+\dfrac{7}{20}=\dfrac{54}{60}\)

yields the puzzle: “Four sons must divide a flock of 54 emus according to the proportions 1/4, 1/6, 2/15, and 7/20.”

Those puzzles that involve unit fractions and the transfer of one sheep for their solution are dubbed particularly appealing. The equations

\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{8}=\dfrac{23}{24}\)

\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{44}=\dfrac{923}{924}\)

\(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}=\dfrac{119}{120}\)

for instance, yield such puzzles.

Exercise: Explain why the equation \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{78}=\dfrac{90}{91}\) does not represent a herd-sharing problem involving the transfer of just one animal. |

Question: A shepherd bequeaths to his four sons the shares 3/5, 1/7, 1/9 and 1/10 of his flock to each son respectively. He has 601 sheep in his flock. Describe a means for the four sons to successfully honor his will. |

** **

Question: Create a “shepherd and her sheep” puzzle based on the equation \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{2}{7}+\dfrac{1}{12}=\dfrac{80}{84}\). How many daughters? How many sheep? |

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