Seventeen Horses/General Problem 2

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Theorem

A man dies, leaving $n$ indivisible and indistinguishable objects to be divided among $m$ heirs.

They are to be distributed in the ratio $\dfrac 1 {a_1} : \dfrac 1 {a_2} : \cdots : \dfrac 1 {a_m}$.

Let $t = \dfrac q r = \ds \sum_{k \mathop = 1}^m \dfrac 1 {a_k}$ expressed in canonical form.

Let $t \ne 1$.

Then it is possible to achieve the required share by adding $s$ objects to the existing $n$ such that:

$s + q = r$

when $q = n$.


This still works whether $q$ is positive or negative.


Proof


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Examples

$31$ Horses

Suppose $31$ horses are to be divided between $3$ sons in the ratio $\dfrac 1 2 : \dfrac 1 3 : \dfrac 1 5$.

We have that:

$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 5 = \dfrac {15 + 10 + 6} {30} = \dfrac {31} {30}$

of which the numerator is seen to equal the total number of horses to be divided.

So we lend $1$ horse to someone, leaving $30$ horses.

The first son takes $\dfrac 1 2$ of these $30$, that is, $15$.

The second son takes $\dfrac 1 3$ of these $30$, that is, $10$.

The third son takes the remaining $5$ horses, then goes and recovers the $1$ horse that was lent, and so has his share of $6$ horses.


$77$ Horses

Suppose $77$ horses are to be divided between $4$ sons in the ratio $\dfrac 1 2 : \dfrac 1 3 : \dfrac 1 4 : \dfrac 1 5$.

We have that:

$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 5 = \dfrac {30 + 20 + 14 + 12} {60} = \dfrac {77} {60}$

of which the numerator is seen to equal the total number of horses to be divided.

So we lend $17$ horses to someone, leaving $60$ horses.

The fourth son takes $\dfrac 1 5$ of these $60$, that is, $12$.

The third son takes $\dfrac 1 4$ of these $60$, that is, $15$.

The second son takes $\dfrac 1 3$ of these $60$, that is, $20$.

The first son takes the remaining $13$ horses, then goes and hunts down the $17$ horses that were lent, and so has his share of $30$ horses.


Sources

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