Henry Ernest Dudeney/Puzzles and Curious Problems/38 - The Picnic/Solution
Puzzles and Curious Problems by Henry Ernest Dudeney: $38$
- The Picnic
- Four married couples had a picnic together, and their refreshments included $32$ bottles of lemonade.
- Mary only disposed of one bottle,
- Anne had two,
- Jane swallowed the contents of three,
- and Elizabeth emptied four bottles.
- The husbands were more thirsty,
- except John MacGregor, who drank the same quantity as his better half.
- Lloyd Jones drank twice as much as his wife,
- William Smith three times as much as his wife,
- and Patrick Dolan four times as much as his wife demanded.
- The puzzle is to find the surnames of the ladies.
- Which man was married to which woman?
Solution
- Mary is the wife of William Smith
- Anne is the wife of Patrick Dolan
- Jane is the wife of John MacGregor
- Elizabeth is the wife of Lloyd Jones.
Proof 1
Let us set up the following system of linear simultaneous equations in matrix form:
- $\begin {pmatrix}
1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\
\end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ d \end {pmatrix} = \begin {pmatrix} 10 \\ 22 \end {pmatrix}$
where:
- $\set {a, b, c, d} = \set {1, 2, 3, 4}$
Conversion to echelon form proceeds as follows:
\(\ds \) | \(\) | \(\ds \paren {\begin {array} {cccc{{|}}c}
1 & 1 & 1 & 1 & 10 \\ 1 & 2 & 3 & 4 & 22 \\ \end {array} }\) |
||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \paren {\begin {array} {cccc{{|}}c}
1 & 1 & 1 & 1 & 10 \\ 0 & 1 & 2 & 3 & 12 \\ \end {array} }\) |
$r_2 \to r_2 - r_1$ | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \paren {\begin {array} {cccc{{|}}c}
1 & 0 & -1 & -2 & -2 \\ 0 & 1 & 2 & 3 & 12 \\ \end {array} }\) |
$r_1 \to r_1 - r_2$ |
The first of these can be rewritten in the form:
- $a = c + 2 d - 2$
from which it is immediate that $a$ and $c$ are of the same parity.
Hence $b$ and $d$ are also of the same parity as each other.
This reduces the possible permutations of $\tuple {a, b, c, d}$ that need to be investigated.
Let $a = 1$, and so $c = 3$:
Then:
\(\ds c + 2 d - 2\) | \(=\) | \(\ds 5\) | when $d = 2$ | |||||||||||
\(\ds c + 2 d - 2\) | \(=\) | \(\ds 9\) | when $d = 4$ |
So $a \ne 1$.
Let $a = 2$, and so $c = 4$:
Then:
\(\ds c + 2 d - 2\) | \(=\) | \(\ds 4\) | when $d = 1$ | |||||||||||
\(\ds c + 2 d - 2\) | \(=\) | \(\ds 8\) | when $d = 3$ |
So $a \ne 2$.
Let $a = 3$, and so $c = 1$:
Then:
\(\ds c + 2 d - 2\) | \(=\) | \(\ds 3\) | when $d = 2$ | |||||||||||
\(\ds c + 2 d - 2\) | \(=\) | \(\ds 7\) | when $d = 4$ |
So $a = 3$, $c = 1$, $d = 2$ is a possible solution, giving $b = 4$.
We check whether $a + 2 b + 3 c + 4 d = 22$, and find:
- $a + 2 b + 3 c + 4 d = 3 + 2 \times 4 + 3 \times 1 + 4 \times 2 = 22$
showing that this is indeed a valid solution.
Let $a = 4$, and so $c = 2$:
Then:
\(\ds c + 2 d - 2\) | \(=\) | \(\ds 2\) | when $d = 1$ | |||||||||||
\(\ds c + 2 d - 2\) | \(=\) | \(\ds 4\) | when $d = 3$ |
So $a = 4$, $c = 2$, $d = 3$ is a possible solution, giving $b = 1$.
We check whether $a + 2 b + 3 c + 4 d = 22$, but find:
- $a + 2 b + 3 c + 4 d = 4 + 2 \times 1 + 3 \times 2 + 4 \times 3 = 24$
showing that this is not a valid solution after all.
So we have eliminated all possible solutions except:
- $\tuple {a, b, c, d} = \tuple {3, 4, 1, 2}$
Proof 2
From:
- $a + 2 b + 3 c + 4 d = 22$
we can infer that:
- $a + 3 c = 22 - 2 b - 4 d$
is even.
Hence $a$ and $c$ must be of the same parity.
Suppose both $a$ and $c$ are even.
Then $b$ and $d$ are both odd.
Either $a = 2$ and $c = 4$ or vice versa.
In either case:
- $a + 3 c \equiv 2 \pmod 4$
Because $b$ will be odd:
- $2 p \equiv 2 \pmod 4$
and so:
- $a + 3 c = 22 - 2 b - 4 d \equiv 0 \pmod 4$
Thus we have a contradiction.
So $a$ and $c$ must both be odd and therefore $b$ and $d$ must both be even.
$d$ cannot be $4$ since any assignment of $\set {1, 2, 3}$ to $\tuple {a, b, c}$ yields:
- $a + 2 b + 3 c > 22 - \paren {4 \times 4} = 6$
So we must have $d = 2$ and $b = 4$, leading to:
- $a + 3 c = 22 - \paren {2 \times 4} - \paren {4 \times 2} = 6$
from which $a = 3$ and $c = 1$ follow immediately.
So:
- Jane drank the same quantity as her husband John MacGregor, that is, $3$ bottles
- Lloyd Jones drank twice as much as the $4$ bottles drunk by his wife Elizabeth
- William Smith drank three times as much as his wife Mary, who drank just $1$ bottle
- Patrick Dolan drank four times as much as his wife Anne's $2$ bottles.
$\blacksquare$
Also see
Sources
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $38$. -- The Picnic