Find the number $abc$ 
The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base 10, where digit $a$ is not zero. Find the three-digit number $abc$.

$$N \equiv abcd \pmod{10000}$$
$$N^2 \equiv abcd \pmod{10000}$$
Then,
$$N(N - 1) \equiv 0 \pmod{10000}$$
Then I get the system:
$$N \equiv 0 \pmod{10000}$$
$$N \equiv 1 \pmod{10000}$$
How to apply CRT or something else?
 A: Your final conclusion is not correct. Note that $10,000$ is not a prime. In fact, since $a\neq 0$, it is impossible.
From
$$N(N-1)=2^4\cdot 5^4k$$
we deduce that either $N$ is a multiple of $2^4$ and $N-1$ is a multiple of $5^4$ or vice versa.
It is possible to handle the problem by brute force, but you may prefer to solve the Bezout identity
$$16x+625y=1$$
A: A more elementary way to solve the problem might be by strategical trial and error.
If you look at the last digit of $N$, for it to be the same as that of $N^2$ it must be one of $\{0,1,5,6\}$, so $d\in \{0,1,5,6\}$. We can discard $0$ since that would mean that $c=0$ wich would also mean that $a = 0$ and we know that not to be the case.

If $d = 1$ then it follows that $c = 0$ (all the other options won't work).
If $d = 5$ then it follows that $c = 2$ (same reason).
If $d = 6$ then if wollows that $c = 7$ (yup, you gessed it).

That means that last two digits of $N$ are $01$, $25$ or $76$.

If they are $01$ then it follows that $b = 0$.
If they are $25$ then it follows that $b = 6$.
If they are $76$ then it follows that $b = 3$.

That means that last three digits of $N$ are $001$, $625$ or $376$.

We can discard the option $001$ since it would mean that $a = 0$.
We can discard the option $625$ since it would mean that $a = 0$.
If they are $376$ then it follows that $a = 9$.

Therefore the last digits of $N$ are $9376$. Indeed $9376^2 = 87909376 \equiv 9376\pmod{10^4}$
Indeed this solution agrees with the comment I made on your question:
$$9376 = 586\cdot 16 \qquad 9375 = 15 \cdot 625$$
Taking this values for $p,\ q,\ k_1,\ k_2$ yield the desired result.
$$p = 16\quad q = 625\quad k_1 = 586\quad k_2 = 15$$
