Henry Ernest Dudeney/Puzzles and Curious Problems/176 - Counting the Loss/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $176$

Counting the Loss
An officer explained that the force to which he belonged originally consisted of $1000$ men, but that it lost heavily in an engagement,
and the survivors surrendered and were marched down to a concentration camp.
On the first day's march one-sixth of the survivors escaped;
on the second day one-eighth of the remainder escaped, and one man died;
on the third day's march one-fourth of the remainder escaped.
Arrived in camp, the rest were set to work in four equal gangs.
How many had been killed in the engagement?


Solution

There were $472$ soldiers who were killed in the engagement.


Proof

Let $n$ be the number of survivors of the engagement.

Let $n_1$, $n_2$ and $n_3$ be the numbers left in the captured group at the end of days $1$ to $3$ respectively.

We have:

\(\ds n_1\) \(=\) \(\ds n - \dfrac n 6 = \dfrac {5 n} 6\) On the first day's march one-sixth of the survivors escaped;
\(\ds n_2\) \(=\) \(\ds n_1 - \dfrac {n_1} 8 - 1 = \dfrac {7 n_1} 8 - 1\) on the second day one-eighth of the remainder escaped, and one man died;
\(\ds n_3\) \(=\) \(\ds n_2 - \dfrac {n_2} 4 = \dfrac {3 n_2} 4\) on the third day's march one-fourth of the remainder escaped.
\(\ds \exists k \in \Z_{>0}: \, \) \(\ds n_3\) \(=\) \(\ds 4 k\) Arrived in camp, the rest were set to work in four equal gangs.
\(\ds \leadsto \ \ \) \(\ds n_2\) \(=\) \(\ds \dfrac {4 n_3} 3\)
\(\ds \) \(=\) \(\ds \dfrac {4 \paren {4 k} } 3\)
\(\ds \) \(=\) \(\ds \dfrac {16 k} 3\)
\(\ds \leadsto \ \ \) \(\ds n_1\) \(=\) \(\ds \dfrac {8 \paren {n_2 + 1} } 7\)
\(\ds \) \(=\) \(\ds \dfrac 8 7 \paren {\dfrac {16 k} 3 + 1}\)
\(\ds \) \(=\) \(\ds \dfrac {8 \paren {16 k + 3} } {21}\)
\(\ds \) \(=\) \(\ds \dfrac {128 k + 24} {21}\)
\(\ds \leadsto \ \ \) \(\ds n\) \(=\) \(\ds \dfrac {6 n_1} 5\)
\(\ds \) \(=\) \(\ds \dfrac 6 5 \paren {\dfrac {128 k + 24} {21} }\)
\(\ds \) \(=\) \(\ds \dfrac {768 k + 144} {105}\)
\(\ds \) \(=\) \(\ds \dfrac {256 k + 48} {35}\)
\(\ds \leadsto \ \ \) \(\ds k\) \(=\) \(\ds \dfrac {35 n - 48} {256}\)

This leads us to the linear Diophantine equation:

$35 n - 256 k = 48$

From Example of Linear Diophantine Equation: $35 x - 256 y = 48$:

$n = 16 + 256 t, k = 2 + 35 t$

for $t \in \Z$.

We now need to select a value of $t$ such that:

$1 \le n \le 1000$

and:

$6 \divides n$

in order to fit the conditions of the question.

By inspection of the cases where $t = 0, 1, 2, 3, 4$ we have that the only solution is for $t = 2$.

Thus:

$n = 16 + 256 \times 2 = 528$

That is, there were $528$ survivors of the initial combat.

Hence $1000 - 528 = 472$ men died in battle.

The result follows.

$\blacksquare$


Sources

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