If a set of markers is placed in rows of $4$ each, there are $2$ markers left over; if in rows of $5$ each, there are $3$ left over; and if in rows of $7$ there are $5$ left over. What is the smallest number of markers that the set could contain?
First I've considered $\operatorname{lcm}(4,5,7)=140$, but couldn't deduce anything from here.
I suppose that there is a simpler way to approach this, but I couldn't figure out anything except the chinese remainder theorem. I have the system of congruences $$\begin{align} x &\equiv 2 \pmod{4} \\ x &\equiv 3 \pmod{5} \\ x &\equiv 5 \pmod{7} \end{align}$$
Using the chinese remainder theorem I've managed to solve this and get that $x \equiv 138 \pmod{140}$. But I'm not sure how to interpret this result in the "applied" way that the problem requires.
I can write $x \equiv 138 \pmod{140}$ as $x=140k +138$ for $k \in \Bbb Z$ and since the problem asks for smallest I have that for $k=0, x=138$ which turns out to be correct, but I don't still quite understand how to interpret the result $x \equiv 138 \pmod{140}$.