Ratio of Number to Reversal which is Multiple

Theorem
Take a (strictly) positive integer $n$, written in conventional decimal notation.

Let $m$ be the reversal of $n$.

Let $m = k n$ where $k$ is an integer.

Then $k$ is either $4$ or $9$.

Existence
$8712 = 4 \times 2178$

$9801 = 9 \times 1089$

Uniqueness
Write $n = \sqbrk {a \dots b}$.

Then $m = \sqbrk {b \dots a} = k n$.

Comparing the leftmost digit, we must have $k a \le b < k \paren {a + 1}$.

Hence $k a < 10$.

Comparing the rightmost digit, we must have $a \equiv k b \pmod {10}$.

We only consider $2 \le k < 10$ since the case $k = 1$ is trivially satisfied by palindromes.

For $k \ge 5$, we must have $a = 1$.

$1 \equiv k b \pmod {10}$ has solutions only for $k = 7$ or $9$.

For $k = 7$, we must have $b = 3$.

However $3 < 7 \times 1$.

Hence $k = 9$.

For $k = 3$, $a \le 3$.

From $a \equiv 3 b \pmod {10}$ we have:
 * $\tuple {a, b} = \tuple {1, 7}, \tuple {2, 4}, \tuple {3, 1}$

However none of the above satisfy the condition:
 * $3 a \le b < 3 \paren {a + 1}$

For $k = 2$, $a \le 4$.

From $a \equiv 2 b \pmod {10}$, both sides must be even.

Hence $a = 2$ or $4$, giving $b = 1$ or $6$ or $2$ or $7$.

However none of the above satisfy the condition:
 * $2 a \le b < 2 \paren {a + 1}$

Hence the only possible values of $k$ are $4$ and $9$.

Also note that

 * $2178 = 2 \times 1089$