The $2n$-digit number divisible by the product of two numbers within itself

Question

I have created this problem on a website (inspired by one of the practice tests I've done)

In case that blue "this problem" button is inaccesible, the problem goes like this:

Find the largest six-digit natural number $\overline{abcdef}$ $(a,d\neq 0)$ that satisfies: $(\overline{abc}\times \overline{def}) \mid \overline{abcdef}$

However I want to generalized it, as the problem below:

Find all numbers $\overline{a_{1}a_{2}a_{3}a_{4}...a_{n}b_{1}b_{2}b_{3}b_{4}...b_{n}}$ with $a_{1},b_{1}\neq 0$ that satisfies: $(\overline{a_{1}a_{2}a_{3}a_{4}...a_{n}}\times \overline{b_{1}b_{2}b_{3}b_{4}...b_{n}}) \mid \overline{a_{1}a_{2}a_{3}a_{4}...a_{n}b_{1}b_{2}b_{3}b_{4}...b_{n}}$

With $n=2$, I found two numbers $1734$ and $1352$ satisfy it.

With $n=3$, I found two numbers $167334$ and $143143$ satisfy it. (And yes, repetition is allowed)

With $n=4$, I found one number $16673334$ satisfy it.

So the question is: Is there any pattern for the numbers above for every integer $n$ and $n>1$?

I have found one number that satisfy it for every integer $n$:

$1666...6673333...334$ with $n-2$ digits are $6$ and $n-1$ digits are $3$, because

$1666...6673333...334 \div (1666...667 \times 3333...334)$

$=(1666...667 \times 10^{n} + 1666...667 \times 2) \div (1666...667 \times 1666...667 \times 2)$

$=\frac{10^n+2}{6}\times \left(10^n+2\right)\div \left(\frac{10^n+2}{6}\times \frac{10^n+2}{6}\times 2\right)$

$=\frac{\left(10^n+2\right)^2}{6}\div \frac{\left(10^n+2\right)^2}{18}=3$

Are there any other "generalized" numbers like above that satisfy the generalized problem above? If there are no more numbers, then how to prove it?

P/S: The math problem that inspires me from making this is from a practice school team test, it requires finding numbers for the case $n=2$. It is quite interesting to find a number with this trait.

Answer 1

if $a$ and $b$ are $n$-digit numbers such that the product $ab$ divides the concatenation $10^n a + b$, then in particular $a \mid 10^n a + b$, so $a \mid b$. Writing $b$ as $ka$, we see that $ka^2 \mid 10^n a + ka$, so $ka \mid 10^n + k$.

Since $k \mid 10^n + k$, we know that $k \mid 10^n$. Since both $a$ and $ka$ are $n$-digit numbers, we know that $k < 10$, so $k \in \{1,2,4,5,8\}$.

Now, since $ka \mid 10^n+k$, write $ka = \frac{10^n+k}{\ell}$. So $b = ka = \frac{10^n+k}{\ell}$ and $a = \frac{10^n+k}{k\ell}$. In particular, since $a \ge 10^{n-1}$, we know that $k\ell \le 10$; since $b < 10^n$, we know $\ell >1$.

(Actually, it's not $k\ell \le 10$ but $k\ell \le \frac{10^n+k}{10^{n-1}} = 10 + \frac{k}{10^{n-1}}$. This only makes a difference when $n=1$; here, the exceptional solutions $11$, $12$, and $15$ violate the constraint $k\ell \le 10$.)

All that remains is to try some cases for $k$ and $\ell$.

• For $k=1$, since any of $\ell = 2, 3, 4, 5, 6, 8, 9, 10$ are never divisors of $10^n+1$, our only hope is setting $\ell=7$, so $a = b = \frac{10^n+1}{7}$. This is an integer when $n \equiv 3 \pmod 6$. For example, when $n=3$ it gives $143143$. When $n=9$ it gives $142857143142857143$. We could go on.
• For $k=2$, $\ell$ must be one of $2,3,4,5$ and divide $\frac{10^n+2}{2}$; the divisibility condition usually rules out $2$, $4$, and $5$. (Exception: when $n=1$, $2$ divides $6$, which leads to $a=3$ and $b=6$.) Taking $\ell=3$ gives us $b = \frac{10^n + 2}{3}$ and $a = \frac{10^n + 2}{6}$, and this is the infinite family of examples already found in the question.
• For $k=4$, $\ell=3$ is already too big (violating $k\ell \le 10$) so $\ell=2$ is the only possibility. However, $\frac{10^n+4}{4}$ is generally not divisible by $2$; the exception is $\frac{104}{4}=26$, giving us $a=13$ and $b=52$.
• For $k=5$, we also need $\ell=2$ to satisfy $k\ell\le 10$, but $\frac{10^n+5}{5} = 2\cdot 10^{n-1}+1$ is never divisible by $2$.
• Finally, for $k=8$, even taking $\ell=2$ violates $k\ell \le 10$, so no examples are to be found here.

As a result, the complete list of solutions is:

• The $143143$ family, including $143143$, $142857143142857143$, $142857142857143142857142857143$, and so on. (We can repeat the $142857$ block any number of times.)
• The $167334$ family, including $24$, $1734$, $167334$, $16673334$, and so on. (We can repeat the $6$ and $3$ any number of times.)
• The exceptional examples $11$, $12$, $15$, $36$ and $1352$ that don't fit into either family.