Henry Ernest Dudeney/Modern Puzzles/82 - Army Figures/Solution

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Modern Puzzles by Henry Ernest Dudeney: $82$

Army Figures
A certain division in an army was composed of a little over twenty thousand men, made up of five brigades.
It was know that one third of the first brigade,
two-sevenths of the second brigade,
seven-twelfths of the third,
nine-thirteenths of the fourth,
and fifteen-twenty-seconds of the fifth brigades happened in every case to be the same number of men.
Can you discover how many men there were in every brigade?


Solution

The $5$ brigades contained respectively $5670$, $6615$, $3240$, $2730$ and $2772$ men.


Proof

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

We are given that:

$\dfrac a 3 = \dfrac {2 b} 7 = \dfrac {7 c} {12} = \dfrac {9 d} {13} = \dfrac {15 e} {22}$

Our first task is to place all these over a common denominator.

To do this we calculate:

$\lcm \set {3, 7, 12, 13, 22} = 3 \times 7 \times 4 \times 13 \times 11 = 12012$

which gives us:

$\dfrac {4004 a} {12012} = \dfrac {3432 b} {12012} = \dfrac {7007 c} {12012} = \dfrac {8316 d} {12012} = \dfrac {8190 e} {12012}$

That is:

$(1) \quad 4004 a = 3432 b = 7007 c = 8316 d = 8190 e$

Now we calculate:

\(\ds \) \(\) \(\ds \lcm \set {4004, 3432, 7007, 8316, 8190}\)
\(\ds \) \(=\) \(\ds \lcm \set {2^2 \times 7 \times 11 \times 13, 2^3 \times 3 \times 11 \times 13, 7^2 \times 11 \times 13, 2^2 \times 3^3 \times 7 \times 11, 2 \times 3^2 \times 5 \times 7 \times 13}\)
\(\ds \) \(=\) \(\ds 2^3 \times 3^3 \times 5 \times 7^2 \times 11 \times 13\)
\(\ds \) \(=\) \(\ds 7 \, 567 \, 560\)

Dividing by each number in turn, we have:

\(\ds \dfrac {7 \, 567 \, 560} {4004}\) \(=\) \(\ds 1890\)
\(\ds \dfrac {7 \, 567 \, 560} {3432}\) \(=\) \(\ds 2205\)
\(\ds \dfrac {7 \, 567 \, 560} {7007}\) \(=\) \(\ds 1080\)
\(\ds \dfrac {7 \, 567 \, 560} {8316}\) \(=\) \(\ds 910\)
\(\ds \dfrac {7 \, 567 \, 560} {8190}\) \(=\) \(\ds 924\)

That is:

\(\ds 7 \, 567 \, 560\) \(=\) \(\ds 1890 \times 4004\)
\(\ds \) \(=\) \(\ds 2205 \times 3432\)
\(\ds \) \(=\) \(\ds 1080 \times 7007\)
\(\ds \) \(=\) \(\ds 910 \times 8316\)
\(\ds \) \(=\) \(\ds 924 \times 8190\)

Thus we have found a set of numbers which fulfil the conditions for $(1)$:

$\set {1890, 2205, 1080, 910, 924}$

and hence so does:

$\set {1890 k, 2205 k, 1080 k, 910 k, 924 k}$

for $k \in \Z_{\ge 1}$.


We have:

$1890 + 2205 + 1080 + 910 + 924 = 7009$

Setting $k = 3$ we get:

$5670 + 6615 + 3240 + 2730 + 2772 = 21027$

which fulfils the condition:

A certain division in an army was composed of a little over twenty thousand men ...

Hence we can state:

\(\ds a\) \(=\) \(\ds 5670\)
\(\ds b\) \(=\) \(\ds 6615\)
\(\ds c\) \(=\) \(\ds 3240\)
\(\ds d\) \(=\) \(\ds 2730\)
\(\ds e\) \(=\) \(\ds 2772\)

$\blacksquare$


Sources

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