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

by : $82$

 * Army Figures

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:

Dividing by each number in turn, we have:

That is:

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: