Henry Ernest Dudeney/Modern Puzzles/8 - Buying Buns/Solution
Modern Puzzles by Henry Ernest Dudeney: $8$
- Buying Buns
- Buns were being sold at three prices:
- 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 Dudeney 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:
\(\ds 2 T \paren {a + \dfrac b 2 + \dfrac c 3}\) | \(=\) | \(\ds 7\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds T \paren {6 a + 3 b + 2 c}\) | \(=\) | \(\ds 21\) | clearing fractions |
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:
- $6 a + 3 b + 2 c = 3$
We note that $b$ must be odd in order to make $3$.
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 $14$ 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 buns.
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 $7 \oldpence$ to buy their buns and share them equitably.
$\blacksquare$
Sources
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $8$. -- Buying Buns
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $5$. Buying Buns