A certain number leaves a remainder of 1 when divided by 3, leaves a remainder of 2 when divided by 4, leaves a remainder of 3 when divided by 5, and leaves a remainder of 4 when divided by 6. What is the smallest positive integer that satisfies these properties?
Solution 1: Let $x$ be the number sought. Observe that $x+2$ is divisible by $3, 4, 5,$ and $6$. The least common multiple of these numbers is $60$. Therefore, $x+2=60k$, so $x=58$.

Solution 2: Let $x$ be the number sought. The remainder of the division of $x$ by $3$ is $1$. Therefore, $x$ is of the form $x=3a+1$, with $a \geq 0$ an integer. Next, we want to determine what form $a$ must take for $x=3a+1$ to leave a remainder of $2$ when divided by $4$, that is, for $3a$ to leave a remainder of $1$ when divided by $4$. What must be the remainder of the division of $a$ by $4$? On the one hand, if this remainder is $3$, then $a$ is of the form $a=4b+3$, from which it follows that $3a=12b+9=4 \cdot (3b+2)+1$ indeed leaves a remainder of $1$ when divided by $4$. On the other hand, we can verify that any other remainder would not work. For example, if $a$ left a remainder of $2$ when divided by $4$, we would have $a=4b'+2$ and $3a=12b'+6=4 \cdot (3b'+1)+2$ would leave a remainder of $2$ and not $1$ when divided by $4$.

Substituting $a=4b+3$ into $x=3a+1$, we get $x=12b+10$. We now use the fact that $x$ leaves a remainder of $3$ when divided by $5$. Since $10$ is a multiple of $5$, $12b$ also leaves a remainder of $3$ when divided by $5$. But $12b=10b+2b$ and $10b$ is a multiple of $5$. Therefore, $2b$ leaves a remainder of $3$ when divided by $5$. Then, $b$ leaves a remainder of $4$ when divided by $5$. Indeed, on the one hand, if $b=5c+4$, then $2b=10c+8=5 \cdot (2c+1)+3$ does indeed leave a remainder of $3$ when divided by $5$. On the other hand, as above, we can verify that $4$ is the only remainder that works. Therefore, $b=5c+4$ and $x=12b+10=12 \cdot (5c+4)+10=60c+58$. We conclude that the solutions to the problem are the numbers $x$ of the form $x=60n+58$, with $n \geq 0$ an integer. The smallest of them, for $n=0$, is $x=58$.