Find an integer $x$ such that $x \equiv 2(\bmod 3), x \equiv 3(\bmod 4)$ and $x \equiv 1(\bmod 5)$.
First, the Chinese Remainder Theorem indicates that such an $x$ does indeed exist, as 3, 4, and 5 are pairwise coprime. Furthermore, it also indicates that there exists a unique natural number $a < 3 \times 4 \times 5 = 60$ such that $x \equiv 2(\bmod 3)$, $x \equiv 3(\bmod 4)$, and $x \equiv 1(\bmod 5)$ if and only if $x \equiv a(\bmod 60)$.

Let us then find such an integer $a$. We know that $a \equiv 2(\bmod 3)$ and that $a \equiv 3(\bmod 4)$. Therefore, we can write $a = 3k + 2$. It is then required that $k$ satisfy the equation $3k + 2 \equiv 3(\bmod 4)$, or equivalently $3k \equiv 1(\bmod 4)$.

Since 3 is invertible modulo 4 and its inverse is 3, this means that $k \equiv 3(\bmod 4)$. We can thus write $k = 4l + 3$, and therefore $a = 12l + 11$.

We then want $l$ to satisfy the equation $12l + 11 \equiv 1(\bmod 5)$, or equivalently $12l \equiv 0(\bmod 5)$. Since 12 is invertible modulo 5, it follows that $l \equiv 0(\bmod 5)$. We can thus write $l = 5m$ and $a = 60m + 11$.

We have thus shown that $a = 11$. In particular, it follows that $x = 11$ satisfies simultaneously the three equations $x \equiv 2(\bmod 3)$, $x \equiv 3(\bmod 4)$, and $x \equiv 1(\bmod 5)$.