Seventeen Horses/General Problem 1

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Theorem

A man dies, leaving $n$ indivisible and indistinguishable objects to be divided among $3$ heirs.

They are to be distributed in the ratio $\dfrac 1 a : \dfrac 1 b : \dfrac 1 c$.

Let $\dfrac 1 a + \dfrac 1 b + \dfrac 1 c < 1$.

Then there are $7$ possible values of $\tuple {n, a, b, c}$ such that the required shares are:

$\dfrac {n + 1} a, \dfrac {n + 1} b, \dfrac {n + 1} c$

These values are:

$\tuple {7, 2, 4, 8}, \tuple {11, 2, 4, 6}, \tuple {11, 2, 3, 12}, \tuple {17, 2, 3, 9}, \tuple {19, 2, 4, 5}, \tuple {23, 2, 3, 8}, \tuple {41, 2, 3, 7}$

leading to shares, respectively, of:

$\tuple {4, 2, 1}, \tuple {6, 3, 2}, \tuple {6, 4, 1}, \tuple {9, 6, 2}, \tuple {10, 5, 4}, \tuple {12, 8, 3}, \tuple {21, 14, 6}$


Proof

It is taken as a condition that $a \ne b \ne c \ne a$.

We have that:

$\dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 n = 1$

and so we need to investigate the solutions to the above equations.


From Sum of 4 Unit Fractions that equals 1, we have that the only possible solutions are:

\(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4\) \(=\) \(\ds 1\)


From these, we can eliminate the following, because it is not the case that $a \ne b \ne c \ne a$:

\(\ds \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4\) \(=\) \(\ds 1\)


Then we can see by inspection that the following are indeed solutions to the problem:

\(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}\) \(=\) \(\ds 1\)
\(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8\) \(=\) \(\ds 1\)


The remaining tuple we have is:

$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15} = 1$

But we note that:

$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} = \dfrac {28} {30}$

which is not in the correct form.

Hence the $7$ possible solutions given.

$\blacksquare$


Sources

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