Find All 4-Digit Kaprekar Numbers | Chinese Remainder Theorem

Description:
Can a four-digit number (n) be split from its square in such a way that the two parts add back to (n)?

In this video, we determine all four-digit positive integers (n) for which there exist positive integers (a) and (b) satisfying
[
n=a+b
\qquad\text{and}\qquad
n^2=10^4a+b.
]

The key step is to transform the conditions into
[
9999\mid n(n-1),
]
and then use the factorisation
[
9999=9\cdot 11\cdot 101.
]

Since consecutive integers are coprime, each of the factors (9), (11), and (101) must divide either (n) or (n-1). This produces eight cases, which can be solved elegantly using modular arithmetic and the Chinese Remainder Theorem.

A fascinating problem connecting digit patterns, divisibility, congruences, and Kaprekar numbers!