Henry Ernest Dudeney/Puzzles and Curious Problems/135 - Distributing Nuts/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $135$

Distributing Nuts
Aunt Martha bought some nuts.
She gave Tommy one nut and a quarter of the remainder;
Bessie then received one nut and a quarter of what were left;
Bob, one nut and a quarter of the remainder;
and, finally, Jessie received one nut and a quarter of the remainder.
It was then noticed that the boys had received exactly $100$ nuts more than the girls.
How many nuts had Aunt Martha retained for her own use?


Solution

Aunt Martha kept $321$ nuts for herself.


Proof

Let $n$ be the total number of nuts.

Let $a$, $b$, $c$, $d$ and $m$ be the number of nuts received or retained by Tommy, Bessie, Bob, Jessie and Aunt Martha respectively.

We have:

\(\text {(1)}: \quad\) \(\ds a + b + c + d + m\) \(=\) \(\ds n\)
\(\text {(2)}: \quad\) \(\ds a\) \(=\) \(\ds 1 + \dfrac {n - 1} 4\) She gave Tommy one nut and a quarter of the remainder;
\(\text {(3)}: \quad\) \(\ds b\) \(=\) \(\ds 1 + \dfrac {n - a - 1} 4\) Bessie then received one nut and a quarter of what were left;
\(\text {(4)}: \quad\) \(\ds c\) \(=\) \(\ds 1 + \dfrac {n - a - b - 1} 4\) Bob, one nut and a quarter of the remainder;
\(\text {(5)}: \quad\) \(\ds d\) \(=\) \(\ds 1 + \dfrac {n - a - b - c - 1} 4\) and, finally, Jessie received one nut and a quarter of the remainder.
\(\text {(6)}: \quad\) \(\ds a + c\) \(=\) \(\ds b + d + 100\) It was then noticed that the boys had received exactly $100$ nuts more than the girls.

Rewriting $(1)$ to $(5)$ in a more convenient form:

\(\text {(1)}: \quad\) \(\ds m\) \(=\) \(\ds n - a - b - c - d\)
\(\text {(2)}: \quad\) \(\ds 4 a\) \(=\) \(\ds n + 3\)
\(\text {(3)}: \quad\) \(\ds 4 b\) \(=\) \(\ds n - a + 3\)
\(\text {(4)}: \quad\) \(\ds 4 c\) \(=\) \(\ds n - a - b + 3\)
\(\text {(5)}: \quad\) \(\ds 4 d\) \(=\) \(\ds n - a - b - c + 3\)
\(\ds \leadsto \ \ \) \(\ds 4 b\) \(=\) \(\ds 3 a\)
\(\ds \leadsto \ \ \) \(\ds 4 c\) \(=\) \(\ds 3 b\)
\(\ds \leadsto \ \ \) \(\ds 4 d\) \(=\) \(\ds 3 c\)
\(\ds \leadsto \ \ \) \(\ds m\) \(=\) \(\ds n - a - b - c - \dfrac {3 c} 4\)
\(\ds \) \(=\) \(\ds n - a - b - \dfrac 7 4 c\)
\(\ds \) \(=\) \(\ds n - a - b - \dfrac 7 4 \times \dfrac 3 4 b\)
\(\ds \) \(=\) \(\ds n - a - \dfrac {37} {16} b\)
\(\ds \) \(=\) \(\ds n - a - \dfrac {37} {16} \dfrac 3 4 a\)
\(\ds \) \(=\) \(\ds n - a - \dfrac {111} {64} a\)
\(\ds \) \(=\) \(\ds n - \dfrac {175} {64} a\)

At this stage we recognise that $a$, $b$, $c$ and $c$ must be positive integers, which gives us the obvious Ansatz:

\(\ds a\) \(=\) \(\ds 64\)
\(\ds b\) \(=\) \(\ds \dfrac 3 4 \times 64 = 48\)
\(\ds c\) \(=\) \(\ds \dfrac 3 4 \times 48 = 36\)
\(\ds d\) \(=\) \(\ds \dfrac 3 4 \times 36 = 27\)

We note that:

$a + c = 100$
$b + d = 75$
\(\ds a + c\) \(=\) \(\ds 100\)
\(\ds b + d\) \(=\) \(\ds 75\)

giving a difference between $a + c$ and $b + d$ of $25$.

So we multiply our Ansatz by $4$ to get:

\(\ds a\) \(=\) \(\ds 256\)
\(\ds b\) \(=\) \(\ds 192\)
\(\ds c\) \(=\) \(\ds 144\)
\(\ds d\) \(=\) \(\ds 108\)

We see that:

\(\ds a + c\) \(=\) \(\ds 400\)
\(\ds b + d\) \(=\) \(\ds 300\)

giving the appropriate difference.

Thus:

\(\ds 4 a\) \(=\) \(\ds n + 3\) recalling $(2)$
\(\ds \leadsto \ \ \) \(\ds n\) \(=\) \(\ds 4 \times 256 - 3\)
\(\ds \) \(=\) \(\ds 1021\)
\(\ds \leadsto \ \ \) \(\ds m\) \(=\) \(\ds 1021 - 256 - 192 - 144 - 108\)
\(\ds \) \(=\) \(\ds 321\)

$\blacksquare$


Sources

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