Find the least value of x which when divided by 3 leaves remainder 1, ... A number when divided by 3 gives a remainder of 1; when divided by 4, gives a remainder of 2; when divided by 5, gives a remainder of 3; and when divided by 6, gives a remainder of 4. Find the smallest such number.
How to solve this question in 1 min?
Any help would be appreciated. :)

Task
Find the smallest $x \in \mathbb{N}_0$ such that:
\begin{align}
x &\equiv 1 \mod 3\\
x &\equiv 2 \mod 4\\
x &\equiv 3 \mod 5\\
x &\equiv 4 \mod 6
\end{align}
Solution
Again, you can solve this quite fast with programming.
The simple solution
Because the numbers are so small that you can use very inefficient solutions (but very fast in terms of programming time):
x = 0
while True:
    if x % 3 == 1 and x % 4 == 2 and x % 5 == 3 and x % 6 == 4:
        print(x)
        break
    x += 1

A faster solution
Every single of the six constraints has to be true. The sixth constraint is only true for every sixth number, so we can "jump" in steps of six:
x = 4
while True:
    if x % 3 == 1 and x % 4 == 2 and x % 5 == 3:
        print(x)
        break
    x += 6

Answer
The answer is 58.
