Modeling Classic Afanti Tales: Donkey, Oil, and Meal Fee Puzzles

I loved Afanti stories as a child and wanted to be as clever as him; mathematical modeling provides methods and frameworks to deepen the understanding of Afanti's wisdom and to continuously enhance problem‑solving abilities.

1 Three Small Stories

1.1 The Donkey Division Story

A family owned 19 donkeys. The deceased owner left a will: the eldest gets one‑half, the second one‑quarter, the third one‑fifth. Since 19 is not divisible by 2, 4, or 5, the division seems impossible. Afanti adds his own donkey, making 20; the eldest receives 10, the second 5, the third 4, and Afanti takes back his donkey.

1.2 The Oil Division Story

Two people bought 10 kg of oil. One has a 7 kg container, the other a 3 kg jar, but they cannot figure out how to split the oil evenly. Afanti suggests repeatedly pouring oil from the bucket into the jar and then from the jar into the container, eventually achieving an equal split.

1.3 The Meal‑Fee Dispute Story

A poor man is sued by a landlord for not paying for a meal he only smelled. Afanti helps the poor man by “paying” with the sound of copper coins, arguing that the landlord heard the sound just as the poor man smelled the aroma, thus no actual payment is required.

2 Three Mathematical Models

2.1 Model for the Donkey Division

According to the will, the three brothers should receive fractions 1/2, 1/4, and 1/5 of the donkeys, which sum to less than 19 and are not integer numbers. Adding Afanti's donkey makes the fractions sum to exactly 19, providing a feasible integer solution.

2.2 Model for the Oil Division

We use a graph‑theoretic approach to solve the oil‑splitting problem.

Define a graph where vertices represent the state (bucket, container, jar) of oil quantities and edges represent possible transfers between containers.

Initial state: (10,0,0). Goal state: (5,5,0). All reachable vertices are enumerated below.

<code>{((0, 7, 3), (3, 7, 0)),((0, 7, 3), (7, 0, 3)),((1, 6, 3), (0, 7, 3)),((1, 6, 3), (1, 7, 2)),((1, 6, 3), (4, 6, 0)),((1, 6, 3), (7, 0, 3)),((1, 7, 2), (0, 7, 3)),((1, 7, 2), (1, 6, 3)),((1, 7, 2), (3, 7, 0)),((1, 7, 2), (8, 0, 2)),((2, 5, 3), (0, 7, 3)),((2, 5, 3), (2, 7, 1)),((2, 5, 3), (5, 5, 0)),((2, 5, 3), (7, 0, 3)),((2, 7, 1), (0, 7, 3)),((2, 7, 1), (2, 5, 3)),((2, 7, 1), (3, 7, 0)),((2, 7, 1), (9, 0, 1)),((3, 4, 3), (0, 7, 3)),((3, 4, 3), (3, 7, 0)),((3, 4, 3), (6, 4, 0)),((3, 4, 3), (7, 0, 3)),((3, 7, 0), (0, 7, 3)),((3, 7, 0), (3, 4, 3)),((3, 7, 0), (10, 0, 0)),((4, 3, 3), (0, 7, 3)),((4, 3, 3), (4, 6, 0)),((4, 3, 3), (7, 0, 3)),((4, 3, 3), (7, 3, 0)),((4, 6, 0), (1, 6, 3)),((4, 6, 0), (3, 7, 0)),((4, 6, 0), (4, 3, 3)),((4, 6, 0), (10, 0, 0)),((5, 2, 3), (0, 7, 3)),((5, 2, 3), (5, 5, 0)),((5, 2, 3), (7, 0, 3)),((5, 2, 3), (8, 2, 0)),((5, 5, 0), (2, 5, 3)),((5, 5, 0), (3, 7, 0)),((5, 5, 0), (5, 2, 3)),((5, 5, 0), (10, 0, 0)),((6, 1, 3), (0, 7, 3)),((6, 1, 3), (6, 4, 0)),((6, 1, 3), (7, 0, 3)),((6, 1, 3), (9, 1, 0)),((6, 4, 0), (3, 4, 3)),((6, 4, 0), (3, 7, 0)),((6, 4, 0), (6, 1, 3)),((6, 4, 0), (10, 0, 0)),((7, 0, 3), (0, 7, 3)),((7, 0, 3), (7, 3, 0)),((7, 0, 3), (10, 0, 0)),((7, 3, 0), (3, 7, 0)),((7, 3, 0), (4, 3, 3)),((7, 3, 0), (7, 0, 3)),((7, 3, 0), (10, 0, 0)),((8, 0, 2), (1, 7, 2)),((8, 0, 2), (7, 0, 3)),((8, 0, 2), (8, 2, 0)),((8, 2, 0), (3, 7, 0)),((8, 2, 0), (5, 2, 3)),((8, 2, 0), (8, 0, 2)),((8, 2, 0), (10, 0, 0)),((9, 0, 1), (2, 7, 1)),((9, 0, 1), (7, 0, 3)),((9, 0, 1), (9, 1, 0)),((9, 0, 1), (10, 0, 0)),((9, 1, 0), (3, 7, 0)),((9, 1, 0), (6, 1, 3)),((9, 1, 0), (9, 0, 1)),((9, 1, 0), (10, 0, 0)),((10, 0, 0), (3, 7, 0)),((10, 0, 0), (7, 0, 3))}</code>

We transform the problem into a shortest‑path search; one feasible edge sequence (oil transfers) is:

Pour oil from the bucket into the jar: (10,0,0) → (7,0,3).

Pour oil from the jar into the container: (7,0,3) → (7,3,0).

Repeat the pour‑jar‑to‑bucket and jar‑to‑container steps as described, eventually reaching (5,5,0), where the oil is evenly split.

Thus the 10 kg of oil is divided equally between the bucket and the container, each holding 5 kg.

2.3 Model for the Meal‑Fee Dispute

To give the story a mathematical form we model “value exchange”. The key idea is:

If one only perceives something without consuming it, its perceived value is zero; if something is consumed or reduced, its value is positive.

Define:

V₁: value of the restaurant’s aroma.

V₂: value of the sound of copper coins.

Assumptions:

The poor man only smelled the aroma, so no food was consumed → V₁ = 0.

The landlord only heard the coin sound, but his wealth did not increase → V₂ = 0.

Logical consequence: the poor man’s “payment” of the coin‑sound value equals the owed aroma value, so no further monetary payment is required.

3 Analysis and Conclusion

From the three stories and their mathematical models we see the power of mathematical modeling to analyze and solve real‑world problems.

3.1 Donkey Division

This illustrates a typical integer‑partition problem; introducing an auxiliary variable (Afanti’s donkey) makes the problem solvable—a common technique in mathematics, computer science, and engineering.

3.2 Oil Division

The story and its model demonstrate how graph theory can describe and solve state‑transition problems.

3.3 Meal‑Fee Dispute

The story reveals the concept of “value exchange”; assigning numerical values to each exchange clarifies analysis and solution.

In summary, mathematical modeling is not only a rigorous discipline but also an art of life, providing fresh perspectives for understanding and tackling practical challenges.