NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Let $N = \overline{a_1a_2\dots a_n b_1b_2\dots b_n}$ be the $2n$-digit number, with $a_1 \ge 1$. Also let $A = \overline{a_1a_2\dots a_n}$ and $B = \overline{b_1b_2\dots b_n}$, so by hypothesis we are given $10^n A + B = N = kAB$ for some positive integer $k$. Multiplying with $k$ and adding $10^n$, we arrive at $(kA - 1)(kB - 10^n) = 10^n$.

For $n = 1$ we immediately solve $(kA-1)(kB-10) = 10$, with $A$ and $B$ not-null digits, to obtain $N \in \{11, 12, 15, 24, 36\}$.

For $n \ge 2$ we have $10^{n-1} \le A < 10^n$ and $1 \le B < 10^n$, thus $1 < k < 10$. A case-by-case discussion follows.

* $k=2$. Then $2 \cdot 10^{n-1} - 1 \le 2A - 1 \mid 10^n$, so $2A - 1 \le 5^n$, forcing $n \le 2$. From $(2A-1)(2B-100) = 100$ we get $N = 1352$.

* $k=3$. Then $3 \cdot 10^{n-1} - 1 \le 3A - 1 \mid 10^n$. We cannot have $3A - 1 = 10^n$, impossible modulo 3. We cannot have $3A - 1 = 2 \cdot 10^{n-1}$, because it is too small. For $3A - 1 = 5 \cdot 10^{n-1}$, we get $A = \frac{5 \cdot 10^{n-1} + 1}{3}$ and $B = 2A = \frac{10^n + 2}{3}$ (for example, for $n=2$ we get $N = 1734$). Thus we get a first infinite family of solutions.

* $k=4$. Then $4 \cdot 10^{n-1} - 1 \le 4A - 1 \mid 10^n$, so $4A - 1 \le 5^n$, impossible.

* $k=5$. Then $5 \cdot 10^{n-1} - 1 \le 5A - 1 \mid 10^n$, so $5A - 1 \le 2^n$, impossible.

* $k=6$. Then $6 \cdot 10^{n-1} - 1 \le 6A - 1 \mid 10^n$, so $6A - 1 \le 10^n$, which forces $6A - 1 = 10^n$, impossible modulo 3.

* $k=7$. Then $7 \cdot 10^{n-1} - 1 \le 7A - 1 \mid 10^n$, so $7A - 1 \le 10^n$, which forces (as in the above) $7A - 1 = 10^n$. This leads to $A = B = \frac{10^n + 1}{7}$, with only proviso that $n$ is an odd multiple of 3 in order to $A, B$ to be integer (for example, for $n=3$ we get $N = 143143$, while for $n=9$ we get $N = 142857143142857143$). Thus we arrive at a second infinite family of solutions.

* $k=8$ and $k=9$ don't work, since they force $kA - 1 = 10^n$, impossible modulo 2 or modulo 3.

Thus, wrapping things up, the solutions are $N \in \{11, 12, 15, 24, 36\}$, the isolated value $N = 1352$, and the two infinite families described in the above.

With the above notations, from $kAB = 10^n A + B$, we get that $10^n A = B(kA-1)$, so $A \mid B(kA-1)$. Since $(A, kA-1) = 1$, we have $A \mid B$, so $B = pA$, where $p \in \mathbb{N}^*$. It follows that $p(kA-1) = 10^n$, so $p \mid 10^n$. As $A$ and $B$ are $n$-digit numbers, we have $1 \le p < 10$, so $p \in \{1, 2, 4, 5, 8\}$.

Moreover, from $p(kA-1) = 10^n$ and $10^{n-1} \le A \le B < 10^n$, we get that $1 + \frac{p}{10^n} < k \le \frac{10}{p} + \frac{1}{10^{n-1}}$, so $2 \le k \le \frac{10}{p} + \frac{1}{10^{n-1}}$. In order to improve this bounding, we'll study separately $n=1$, in which case we easily find the solutions $N \in \{11, 12, 15, 24, 36\}$.

For $n \ge 2$, it follows that we have $2 \le k \le \frac{10}{p}$, which immediately drops the case $p=8$. Also, for $p \in \{4, 5\}$, we must have $k=2$. If $p=5$, then $2A - 1 = 5 \cdot 10^{n-1}$, impossible, for parity reasons. If $p = 4$, we get $2A - 1 = 25 \cdot 10^{n-2}$. Again, parity arguments provide $n = 2$, so $A = 13$ and $B = 52$, so $N = 1352$. That leaves us with $p \in \{1, 2, 4\}$.

For $p = 1$, we have $2 \le k \le 10$ and $kA = 10^n + 1$, so $k$ must be odd, and it cannot be 5 or divisible by 3. The only possibility remains $k = 7$, which works only if $7 \mid 10^n + 1$; we get that $n = 6t + 3$, $t \in \mathbb{N}$, so
$$A = B = \frac{10^{6t+3} + 1}{7} \quad \text{and} \quad N = \underbrace{142\,857 \dots 142\,857}_{t \text{ times } 142857} 143.$$

For $p = 2$, we have $2 \le k \le 5$ and $kA = 5 \cdot 10^{n-1} + 1$, so $k$ is odd and it is not 5, so it must be 3 (which works for all $n \ge 1$). It follows that
$$A = \frac{5 \cdot 10^{n-1} + 1}{3}, \quad B = \frac{10^n + 2}{3}, \quad N = \underbrace{1\,66\dots6\,7\,33\dots3\,4}_{n-1 \text{ times } 6}, \quad \underbrace{\text{n times 3}}$$