Henry Ernest Dudeney/Modern Puzzles/8 - Buying Buns

by : $8$

 * Buying Buns


 * Buns were being sold at three prices:
 * one a penny,
 * two a penny,
 * and three a penny.
 * Some children (there were as many boys as girls) were given sevenpence to spend on these buns, each receiving exactly alike.


 * How many buns did each receive?


 * Of course no buns were divided.

Solution
The solution given by is:


 * There must have been three boys and three girls,
 * each of whom received two buns at three a penny
 * and one bun at two a penny,
 * the cost of which would be exactly sevenpence.

However, there are in fact $12$ solutions.

Let them be presented in the form $\tuple {a, b, c}$ where $a$, $b$ and $c$ are the numbers of penny buns, two-for-a-penny buns and three-for-a-penny buns received by each child:


 * $14$ children in total:
 * $\tuple {0, 1, 0}$


 * $6$ children in total (the solution given):
 * $\tuple {0, 1, 2}$


 * $2$ children in total:
 * $\tuple {0, 1, 9}$
 * $\tuple {0, 3, 6}$
 * $\tuple {0, 5, 3}$
 * $\tuple {0, 7, 0}$


 * $\tuple {1, 1, 6}$
 * $\tuple {1, 3, 3}$
 * $\tuple {1, 5, 0}$


 * $\tuple {2, 1, 3}$
 * $\tuple {2, 3, 0}$


 * $\tuple {3, 1, 0}$

Proof
Let $2 T$ be the total number of children.

Let $a$ be the number of penny buns bought by each child.

Let $b$ be the number of two-for-a-penny buns bought by each child.

Let $c$ be the number of three-for-a-penny buns bought by each child.

All of $a$, $b$, $c$ and $T$ are natural numbers where $T > 0$.

Hence we have:

Thus $T$ is a divisor of $21$, and so can be either $1$, $3$, $7$ or $21$.

$21$ is clearly off the table, because that would give:
 * $21 \paren {6 a + 3 b + 2 c} = 21$

and so:
 * $6 a + 3 b + 2 c = 1$

which cannot be satisfied in natural numbers.

Let $T = 7$.

Then we have:


 * $\paren 6 a + 3 b + 2 c = 3$

We note that $b$ must be odd in order to make $7$.

So whatever happens, $b \ge 1$.

When $b = 1$ we have:
 * $6 a + 3 \times 1 + 2 c = 3$

allowing the solution:
 * $\tuple {a, b, c} = \tuple {0, 1, 0}$

That is all you get for $T = 7$.

So, there can be $7$ children each receiving a single two-for-a-penny bun.

That is, $14$ buns are bought, and each child receives one.

Let $T = 3$.

Then we have:


 * $6 a + 3 b + 2 c = 7$

Setting $b = 1$ again gives:
 * $6 a + 3 \times 1 + 2 c = 7$

This leads to:
 * $6 a + 2 c = 4$

for which the only solution is $c = 2$.

This allows the solution:
 * $\tuple {a, b, c} = \tuple {0, 1, 2}$

So there can be $6$ children each receiving $1$ two-for-a-penny bun, and $2$ three-for-a-penny bun.

Finally let $T = 1$.

This leads to:


 * $6 a + 3 b + 2 c = 21$

which gives rise to the solutions for $\tuple {a, b, c}$ as:


 * $\tuple {0, 1, 9}$
 * $\tuple {0, 3, 6}$
 * $\tuple {0, 5, 3}$
 * $\tuple {0, 7, 0}$


 * $\tuple {1, 1, 6}$
 * $\tuple {1, 3, 3}$
 * $\tuple {1, 5, 0}$


 * $\tuple {2, 1, 3}$
 * $\tuple {2, 3, 0}$


 * $\tuple {3, 1, 0}$

So $2$ children have a total of $10$ different ways they can pool their $7 \, \mathrm d.$ to buy their buns and share them equitably.