Find all integers congruent simultaneously to 1 modulo 2, to 2 modulo 3, to 3 modulo 5, and to 4 modulo 7.
This is an immediate practical application of the Chinese theorem: 2, 3, 5, 7 being pairwise coprime, since they are prime, any system of equations: $x \equiv a$ $(\bmod 2), x \equiv b(\bmod 3), x \equiv c(\bmod 5), x \equiv d(\bmod 7)$ admits an infinite number of solutions, of the form: $x \equiv N(\bmod 2 \times 3 \times 5 \times 7)$. We start by solving the last two equations: $x \equiv 4(\bmod 7)$ leads to $x \equiv 4,11,18,25$ or $32(\bmod 5 \times 7)$. Since 5 and 7 are coprime, $4,11,18,25,32$ take all possible values modulo 5, so once the desired value $3: x \equiv 3(\bmod 5)$ and $x \equiv 4(\bmod 7)$ leads to $x \equiv 18(\bmod 35)$. Then $x \equiv 18,53$ or 88 $(\bmod 3 \times 35)$. Among these three values, one and only one, 53, is congruent to 2 modulo 3. Therefore,
$x \equiv 53$ or $203(\bmod 2 \times 105))$. One and only one of these two values is congruent to 1 modulo 2, in this case 53. Thus, the solution to the problem is: $x \equiv 53(\bmod 210)$.