Henry Ernest Dudeney/Puzzles and Curious Problems/134 - The Bag of Nuts/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $134$

The Bag of Nuts
There are $100$ nuts distributed between $5$ bags.
In the first and second there are altogether $52$ nuts;
in the second and third there are $43$;
in the third and fourth there are $34$;
in the fourth and fifth, $30$.
How many nuts are there in each bag?


Solution

In the first to fifth bags in order, the number of nuts is:

$27, 25, 18, 16, 14$


Proof

Let $a$, $b$, $c$, $d$ and $e$ be the number of nuts in each of the first, second, third, fourth and fifth bags respectively.

We have:

\(\ds a + b + c + d + e\) \(=\) \(\ds 100\) There are $100$ nuts distributed between $5$ bags.
\(\ds a + b\) \(=\) \(\ds 52\) In the first and second there are altogether $52$ nuts;
\(\ds b + c\) \(=\) \(\ds 43\) in the second and third there are $43$;
\(\ds c + d\) \(=\) \(\ds 34\) in the third and fourth there are $34$;
\(\ds d + e\) \(=\) \(\ds 43\) in the fourth and fifth, $30$.


We set up this system of linear simultaneous equations in matrix form as:

$\begin {pmatrix}
1 &  1 &  1 &  1 &  1 \\
1 &  1 &  0 &  0 &  0 \\
0 &  1 &  1 &  0 &  0 \\
0 &  0 &  1 &  1 &  0 \\
0 &  0 &  0 &  1 &  1 \\

\end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ d \\ e \end {pmatrix} = \begin {pmatrix} 100 \\ 52 \\ 43 \\ 34 \\ 30 \end {pmatrix}$

It remains to solve this matrix equation.


In reduced echelon form, this gives:

$\begin {pmatrix}
1 &  0 &  0 &  0 &  0 \\
0 &  1 &  0 &  0 &  0 \\
0 &  0 &  1 &  0 &  0 \\
0 &  0 &  0 &  1 &  0 \\
0 &  0 &  0 &  0 &  1 \end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ d \\ e \end {pmatrix} = \begin {pmatrix} 27 \\ 25 \\ 18 \\ 16 \\ 14 \end {pmatrix}$


from which the contents of the bags can be read off directly.

$\blacksquare$


Sources

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