Find smallest $x$ such that $x=59 \pmod {60}$ and $x=1 \pmod 7$ Is a simple way to solve the problem? The method I used is to list all numbers from equation (1) and then see which one give remainder $1$ when divided by $7$. This doesn't seems a very smart way.

Problem: Find smallest $x$ that satisfies
  $$x=59 \pmod {60} \tag 1$$ $$x=1 \pmod 7 \tag 2$$

 A: We can look at $59$, $59+60$, $59+120$, and so on until we bump into something congruent to $1$ modulo $7$. Doesn't take long!
For a more "general" approach, we want to find $k$ so that $59+60k\equiv 1\pmod{7}$. Reducing mod $7$ we find that we want 
$$4k\equiv -58\equiv -2\pmod{7}.$$ 
Multiply through by the inverse of $4$ modulo $7$, which is $2$. We get $k\equiv -4\equiv 3\pmod{7}$. So $59+(3)(60)$ is a solution.
A: Find $a,b$ such that $60a+7b=1$. Use the extended Euclidean algorithm if necessary.
Then $x = 1\cdot60a + 59\cdot7b$ is a solution. All solutions are congruent to this one mod $60\cdot 7$.
A: In this case you can use some heuristics.  Solutions to (1) are of the form $59 + 60n$, $n \in \mathbb{Z}$, so all the positive ones end in $9$.
The first solution to (2) that ends in $9$ is $29$, and every tenth solution thereafter also ends in $9$ (the next one being $99$).
Pretty quickly you can arrive at $239$ as the smallest positive answer.
A: The the technique of "noodling around."
$7*8 = 56 \equiv -4(\mod 60)\\
7*9 = 63 \equiv 3(\mod 60)\\
1 + -2 \equiv 59(\mod 60)\\
1 + 7*(9+9+8+8)\equiv 1 - 2 \equiv 59(\mod 60)\\
239$
Now this is one solution.  Since 7 and 60 are co-prime, $\text{lcm(7,60)} = 420$
$239+420k$
and 239 is the smallest positive solution.
