Henry Ernest Dudeney/Puzzles and Curious Problems/38 - The Picnic/Solution/Proof 1

Proof
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:

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:

So $a \ne 1$.

Let $a = 2$, and so $c = 4$:

Then:

So $a \ne 2$.

Let $a = 3$, and so $c = 1$:

Then:

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:

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:


 * $\tuple {a, b, c, d} = \tuple {3, 4, 1, 2}$

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.