Train Riddle

Let $S$ be the total number of seats in the train and $x$ be the number of seats in compartment $X$ , $y$ be that in each $Y$ , and $z$ be that in each $Z$ .

Then we can set up the Diophantine equations as followed:

$7x + 5y = S-1$ $8y + 3z = S-1$ $4z + 9x = S-1$

We can then rewrite the whole system as matrix multiplication:

$\begin{bmatrix}{7} && {5} && {0} \\ {0} && {8} && {3} \\ {9} && {0} && {4}\end{bmatrix} \begin{bmatrix}{x}\\{y}\\{z}\end{bmatrix} = \begin{bmatrix}{S-1}\\{S-1}\\{S-1}\end{bmatrix} = (S-1)\begin{bmatrix}{1}\\{1}\\{1}\end{bmatrix}$

The determinant of the $3\times3$ matrix equals to $7\times 8 \times 4 + 5\times 3\times 9 = 224+135 = 359$

Hence, solving the equation by multiplying the inverse matrix, we will obtain:

$\begin{bmatrix}{x}\\{y}\\{z}\end{bmatrix} = \dfrac{S-1}{359} \begin{bmatrix}{8\times 4} && {-5\times 4} && {5\times 3} \\ {-9\times -3} && {7\times 4} && {-7\times 3} \\ {8\times -9} && {-5\times -9} && {7\times 8}\end{bmatrix} \begin{bmatrix}{1}\\{1}\\{1}\end{bmatrix} = \dfrac{S-1}{359}\begin{bmatrix}{32-20+15}\\{27+28-21}\\{-72+45+56}\end{bmatrix} = \dfrac{S-1}{359}\begin{bmatrix}{27}\\{34}\\{29}\end{bmatrix}$

To yield integer solutions for $x,y,z$ , $S-1$ must be multiple of $359$ . Then let $S-1 = 359n$ for some positive integer $n$ . Then $x = 27n$ ; $y = 34n$ ; and $z = 29n$ .

We know that $S$ is a multiple of $x+y+z$ as stated in the question, so we can rewrite $S = m(x+y+z) = m(27n+34n+29n) = 90mn$ for some positive integer $m$ .

Hence, we can set up new equation:

$S - 1 = 90mn -1 = 359n$

$90mn-359n = n(90m - 359) = 1$

Clearly, only $n = 1$ can divide $1$ , leading to $m=4$ and $S = 4\times 90 =360$ . Finally, $x = 27$ ; $y = 34$ ; and $z = 29$ . Then $x+y+z = 90$ .