Thailand Mathematical Olympiad

We first show that for $n \in \mathbb{N}$, the integer $n^3-1$ is divisible by $2556 = 2^2 \cdot 3^2 \cdot 71$ if and only if it is of the form $n = 852k + 1$ ($k \in \mathbb{N}$).
(⇒) If $2556 \mid (n^3 - 1)$, then
$$n^3 \equiv 1 \pmod{2^2 \cdot 3^2 \cdot 71} \quad (1)$$
and so $n^3 \equiv 1 \pmod{71}$. Since $71 \nmid n$, Fermat's little theorem implies that $n^{70} \equiv 1 \pmod{71}$, thus
$$1 \equiv n \cdot n^{69} \equiv n(n^3)^{23} \equiv n(1)^{23} \equiv n \pmod{71},$$
i.e.,
$$71 \mid (n-1). \quad (2)$$
From (1), we see that $4 \mid n^3 - 1 = (n-1)(n^2 + n + 1)$. Since $n^2 + n + 1 = n(n+1) + 1$ is odd, we have
$$4 \mid (n-1). \quad (3)$$
If $3 \nmid (n-1)$, then $3 \mid n(n+1)$, so that $3 \nmid n(n+1) + 1 = n^2 + n + 1$, yielding $3 \nmid n^3 - 1$, contradicting (1). Thus,
$$3 \mid (n-1). \quad (4)$$
The three divisibility conditions (2)-(4) show that $n = (3 \times 4 \times 71 \times k)+1 = 852k + 1$ for some $k \in \mathbb{N}$.
(↔) If $n$ is of the form $n = 852k + 1$ ($k \in \mathbb{N}$), then
$$n^3 - 1 = (n-1)(n(n-1) + 2(n-1) + 3) = 2556k(284nk + 568k + 1).$$
To solve the problem, we must determine a five-digit $n$ of the form $852k+1$ with least digit sum. Since $852k$ does not end with 9, it suffices to find an integer of the form $852k$ with least digit sum. Since $4 \mid 852k$, the integer formed from its last two digits are also divisible by 4. To find the integer with least digit sum, we observe that a five-digit integer of the form $abc00$ is divisible by 4 and has two 0 digits making it a good candidate. Since 3 and 71 are factors of the required integer and since 100 is not divisible by 3 nor by 71, we need $abc$ to be divisible by $3 \times 71 = 213$, which has digit sum 6. We anticipate that 21300 is the sought after integer. To verify this, consider other integers with digit sums < 6 and last two digits divisible by 4. All of them are $abc04, abc12, abc20, abc32, abc40$. Since the required integer is divisible by 3, so are their digit sums, implying that the only possible integer is $abc20$, and so $a, b, c \in \{0, 1\}$, $a \neq 0$ and $a+b+c=1$. The only possible integer is 10020. But, this integer is not divisible by 71. Hence, the sought after integer is $21300+1=21301$.