A bus carries $67$ travelers of three types 
A bus carries $67$ travelers of three types:

*

*travelers who pay the entire ticket, which costs $\$3200$.


*students who have a $43$% discount.


*local retirees who only pay $23$% of the ticket price.
The bus collection on that trip was $\$6,292,000$. Calculate the number of travelers in each class knowing that the number of retirees was the same as the number of other travelers.

Solution:

*

*$x$: travelers who pay the entire ticket, $\$3200$


*$y$: students who have a $43\%$ discount, that is, they pay the $57\%$ of the total ticket. ($3200*57\%=1824$)


*$z$: local retirees who only pay $23$% of the ticket price. ($3200*23\%=736$)
The equations are
$$\begin{aligned}
x + y +z &= 67 \\
100x + 57y +23z &= 196625 \\
x+y-z &= 0
\nonumber
\end{aligned}$$
The solutions are $x =  \frac{193945}{43}$, $y = -\frac{385009}{86} $ and $z = 33.5$. However, $y$ is negative. If they ask about the number of travelers, what does it tell me in the result?

equations:
1) $x+y+z=0$
2) $3200x+1824y+736z=6292000$
3) $z=x+y$
 A: You wrote "Yes, I am sure that the exercise data are correct" but it cannot be.
Let $n$ be the number of "normal" travelers, $s$ the number of students and $r$ the number of retirees and put numbers to have
$$ 3200 n+1824 s+736 r=T\tag 1$$
$$n+s+r=P\tag 2$$
$$r=n+s\tag 3$$ where $T$ is the bus collection and $P$ the number of passengers.
Using $(2)$ and $(3)$ makes $r=\frac P 2$ so $P$ cannot be odd.
Second remark : if everyone pays the full price, it would be $3200P$ and $67\times 3200=214400$ which does not have anything to do with the $6292000$ given. To get such a collection, the bus would be bigger than a train !
Just solving the three equations would give
$$n=\frac{T-1280 P}{1376}\qquad s=\frac{1968P-T}{1376}\qquad r=\frac P2$$ and each of $n,s,r$ must be an integer (again with $T \leq 3200 P$).
A: Following Claude Leibovici's answer,
we have $$s-n=\frac{3248r - T}{688}.$$
Noting that $32\mid T,$ let $T=32k,$ then 
$$s-n=\frac{203r - 2k}{43}.$$
If we set $T=6292000$ as in the original problem, then $k = 196625.$
Integer solutions
occur for $r = 43m + 1966$ and $s-n = 203m + 136$
where $m$ is an integer. 
But we also have $s+n=r,$
so $$s = \tfrac12(r + (s-n)) = 123 m + 1051$$
and $$n = \tfrac12(r - (s-n)) = 915 - 80 m = 5 (183 - 16 m).$$
Since $n$ and $s$ must be non-negative, this limits the possible solutions to
$-8\leq m \leq 11.$
Hence the minimum number of retirees occurs when $m = -8$ and 
$r = 43(-8)+1966 = 1622.$
There are then $67$ students and $1555$ other passengers.
The problem is therefore solvable if $67$ is the number of students
rather than the total number of passengers.
