Rhind Papyrus 40
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Problem
- A hundred loaves to five men, one-seventh of the three first men to the two last.
Solution
This means:
- Divide $100$ loaves between $5$ men so that:
- the shares are in arithmetic progression
- the sum of the two smaller shares is $\dfrac 1 7$ of the sum of the $3$ greatest.
Thus, let:
- $a$ be the smallest share, that is, the initial term
- $d$ be the common difference.
We have:
\(\text {(1)}: \quad\) | \(\ds 5 \paren {a + \frac {5 - 1} 2 d}\) | \(=\) | \(\ds 100\) | Sum of Arithmetic Sequence | ||||||||||
\(\text {(2)}: \quad\) | \(\ds a + \paren {a + d}\) | \(=\) | \(\ds \dfrac 1 7 \paren {\paren {a + 2 d} + \paren {a + 3 d} + \paren {a + 4 d} }\) | the second condition | ||||||||||
\(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds a + 2 d\) | \(=\) | \(\ds 20\) | from $(1)$ and simplifying | |||||||||
\(\text {(4)}: \quad\) | \(\ds 11 a - 2 d\) | \(=\) | \(\ds 0\) | from $(2)$ and simplifying | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 12 a\) | \(=\) | \(\ds 20\) | adding $(3)$ to $(4)$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 5 3\) | simplifying | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 d\) | \(=\) | \(\ds 20 - \dfrac 5 3\) | substituting for $a$ in $1$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds d\) | \(=\) | \(\ds \dfrac {55} 6\) | simplifying |
So the shares are:
\(\ds \dfrac 5 3\) | \(=\) | \(\ds 1 \dfrac 2 3\) | ||||||||||||
\(\ds \dfrac 5 3 + \dfrac {55} 6\) | \(=\) | \(\ds 10 + \dfrac 2 3 + \dfrac 1 6\) | ||||||||||||
\(\ds \dfrac 5 3 + 2 \times \dfrac {55} 6\) | \(=\) | \(\ds 20\) | ||||||||||||
\(\ds \dfrac 5 3 + 3 \times \dfrac {55} 6\) | \(=\) | \(\ds 29 + \dfrac 1 6\) | ||||||||||||
\(\ds \dfrac 5 3 + 4 \times \dfrac {55} 6\) | \(=\) | \(\ds 38 + \dfrac 1 3\) |
$\blacksquare$
Sources
- c. 1650 BCE: Ahmes: Rhind Papyrus: Problem $30$
- 1923: T. Eric Peet: The Rhind Mathematical Papyrus: Problem $30$
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Sharing the Loaves: $10$