Remainder via the Chinese remainder theorem

Statement

Find the smallest positive integer $x$ leaving remainder $1$ on division by $3$, remainder $2$ on division by $5$, and remainder $3$ on division by $7$.

Solution

We seek $x\equiv 1\ (3)$, $x\equiv 2\ (5)$, $x\equiv 3\ (7)$. From the first two: numbers $\equiv 2\pmod 5$ are $2,7,12,\dots$; the first $\equiv 1\pmod 3$ is $7$, so $x\equiv 7\pmod{15}$. Adding the third: $7,22,37,52$ have remainders $0,1,2,3$ modulo $7$; hence $52$ works. The smallest positive integer is $x=\boxed{52}\qquad(\,52=3\cdot 17+1=5\cdot 10+2=7\cdot 7+3\,).$