# Rhind Papyrus 40

## 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:

 $\ds \dfrac 5 3$ $=$ $\ds 1 \dfrac 2 3$ $\ds \dfrac 5 3 + \dfrac {55} 6$ $=$ $\ds 10 + \dfrac 2 3 + \dfrac 1 6$ $\ds \dfrac 5 3 + 2 \times \dfrac {55} 6$ $=$ $\ds 20$ $\ds \dfrac 5 3 + 3 \times \dfrac {55} 6$ $=$ $\ds 29 + \dfrac 1 6$ $\ds \dfrac 5 3 + 4 \times \dfrac {55} 6$ $=$ $\ds 38 + \dfrac 1 3$

$\blacksquare$