I have a collection of cards, then:
- If I put them in stacks of 2, I have 1 left
- If I put them in stacks of 3, I have 1 left
- If I put them in stacks of 4, I have 1 left
- If I put them in stacks of 7, I have 0 left
So how many cards do I have??
I have a collection of cards, then:
So how many cards do I have??
Let the number of cards be $x$. Then you would set up the system of linear congruences as follows:
$x \equiv 1 \ (mod \ 3)$
$x \equiv 1 \ (mod \ 4)$
$x \equiv 0 \ (mod \ 7)$
[Note that the condition of $x \equiv 1 \ (mod \ 2)$ intuitively means that $x$ is odd. This information is already completely contained in the congruence modulo 4.]
[Note also that 3, 4 and 7 are pairwise coprime integers.]
Would you know how to solve this system?
As $x \equiv 1 \ (mod \ 3)$, and $x$ is positive, $x$ could be 1, 4, 7, 10, ... i.e. of the form $3k+1$ where $k$ is a positive integer.
Then you look at which of these possibilities for $x$ are congruent to 1 mod 4, etc.
See CRT to solve problems like these.
Your first condition is redundant given the third condition.
So you want a number $x$ such that it is of the form $12k+1$ and it should be divisible by $7$. Trying a few values of $k$, we see $49$ works.