Find all integers $n$ satisfying: $n \equiv 3(\bmod 5), n \equiv 4(\bmod 7), n \equiv 5(\bmod 3)$.
$n=5 k_{1}+3$ must be congruent to 4 modulo 7, hence $5 k_{1} \equiv 1(\bmod 7)$. This is satisfied if $k_{1} \equiv 3$ $(\bmod 7)$, in which case $: n=5 k_{1}+3=35 k_{2}+18$, and this must be congruent to 5 modulo 3 (we could have written $\equiv 2$ or $-1(\bmod 3)$, it would have been the same). In any case, $35 k_{2}$ must be congruent to $5-18=-13(\bmod 3)$, so $2 k_{2} \equiv 2(\bmod 3): k_{2}=3 k_{3}+1$, hence $n=5 k_{1}+3=$ $35 k_{2}+18=105 k_{3}+53$. This is the general solution of the system: any $n$ of the form $105 k+53$
satisfies the three congruences of the statement, and according to the reasoning above, they alone are solutions.