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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??

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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.

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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.

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