Rhind Papyrus 40

From ProofWiki
Jump to navigation Jump to search

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