Consider the number $$2058$$. It is a reversible factor if the number formed by writing it in reverse $$(8502)$$ is a multiple. Likewise, a number is a reversible multiple if the number formed by writing it in reverse is a factor. Of course, $$2058$$ is not reversible because $$8502 \div 2058 = 4.131$$, which is not a whole number. The question immediately arises: Are there reversible factors? If so, what is the smallest reversible factor?

Let's begin with a few preliminary considerations. First no single digit numbers can be reversible, for a single digit written in reverse is the same number. So the smallest reversible number must have at least two digits. Second, the factor by which you multiply a reversible factor to get a reversible multiple must be a single digit between $$2$$ and $$9$$. Zero, of course is out of the question. No number but $$0$$ can have $$0$$ for a factor. Likewise, $$1$$ is completely uninteresting since it is a factor of every number. So the number $$11$$, which might otherwise be considered the smallest reversible factor $$(11 = 11 \times 1)$$ is excluded. Of course, any number greater than $$9$$ will yield a multiple with more digits than the factor. For example, $$11 \times 10 = 110$$ might work if we were to allow leading zeros (e. g., $$011$$), but in fact $$11$$ is a two-digit number while $$110$$ is a three-digit number. A related consequence is that no reversible number may begin or end with $$0$$. Now let's take a look at the factors $$2$$ through $$9$$ in turn and see if any will yield reversible numbers.

A reversible number multiplied by $$2$$ must begin with a $$1$$, $$2$$, $$3$$, or $$4$$. Any higher digit will cause the resulting number to have an extra digit. Likewise, it must end in $$2$$, $$4$$, $$6$$, or $$8$$. So we have the following possibilities: $$1 \dots 2$$, $$2 \dots 4$$, $$3 \dots 6$$, and $$4 \dots 8$$. Multiplying each of these by $$2$$ gives: $$2 \dots 4$$ (not $$2 \dots 1$$), $$4 \dots 8$$ (not $$4 \dots 2$$), $$6 \dots 2$$ (not $$6 \dots 3$$), and $$8 \dots 6$$ (not $$8 \dots 4$$). Clearly, we cannot get a reversible number using $$2$$ as a factor.

A reversible number multiplied by $$3$$ must begin with $$1$$, $$2$$, or $$3$$. Four and greater will result in too many digits. Likewise it must end in $$3$$, $$6$$, or $$9$$. So we have the following possibilities: $$1 \dots 3$$, $$2 \dots 6$$, and $$3 \dots 9$$. Multiplying each by $$3$$ gives: $$3 \dots 9$$ (not $$3 \dots 1$$), $$6 \dots 8$$ (not $$6 \dots 2$$), and $$9 \dots 7$$ (not $$9 \dots 3$$). Again, no reversible numbers with $$3$$ as a multiplier.

A reversible number multiplied by 4 must begin with $$1$$ or $$2$$. Three and greater will cause too many digits. Likewise it must end in $$4$$ or $$8$$, thus: $$1 \dots 4$$, and $$2 \dots 8$$. Multiplying each by $$4$$ gives: $$4 \dots 6$$ (not $$4 \dots 1$$) and $$8 \dots 2$$, which happens to work fine! So our first candidate is $$2 \dots 8$$. Let's move on.

A reversible number multiplied by $$5$$ or greater must begin with $$1$$ and end with the multiplier. This yields the following: $$1 \dots 5 \times 5 = 5 \dots 5$$ (not $$5 \dots 1$$), $$1 \dots 6 \times 6 = 6 \dots 6$$ (not $$6 \dots 1$$), $$1 \dots 7 \times 7 = 7 \dots 9$$ (not $$7 \dots 1$$), $$1 \dots 8 \times 8 = 8 \dots 4$$ (not $$8 \dots 1$$), and $$1 \dots 9 \times 9 = 9 \dots 1$$, which is promising.

We now have two candidates: $$2 \dots 8 \times 4 = 8 \dots 2$$ and $$1 \dots 9 \times 9 = 9 \dots 1$$. No two-digit solutions are possible since $$28 \times 4 \ne 82$$ and $$19 \times 9 \ne 91$$. Consider the three-digit possibilities: $$2a8 \times 4 = 8a2$$ and $$1a9 \times 9 = 9a1$$. Since no carry is allowed when multiplying the second digit, we get $$4a + 3 = a$$ and $$9a + 8 = a$$. Both equations yield $$a = -1$$, which is clearly impossible. So there are no three-digit solutions.

How about 4-digits? Our candidates are $$2ab8 \times 4 = 8ba2$$ and $$1ab9 \times 9 = 9ba1$$. So $$4 \times b + 3$$ ends in $$a$$, and the carry when added to $$4 \times a = b$$ with no carry. Therefore, $$a \le 2$$. Furthermore, $$a$$ must be odd because $$4 \times b + 3$$ ends in $$a$$, and you can’t add 3 to an even number and get another even number. Therefore $$a = 1$$ is the only possibility. If $$4 \times b + 3$$ ends in $$1$$, then $$4 \times b$$ ends in $$8$$. The only multiples of $$4$$ that end in $$8$$ are $$8$$ and $$28$$, so $$b$$ must be $$2$$ or $$7$$. Of these $$7$$ is the only one that works. $$2178 \times 4 = 8712$$. Similar reasoning yields $$1089 \times 9 = 9801$$.

Therefore, the smallest reversible factor is $$1089$$. The only other 4-digit reversible factor is $$2178$$, which happens to be $$2 \times 1089$$. Not only that, all four numbers are multiples of the smallest palindromic number, $$11$$—twice. ($$1089 = 11^2 \times 3^2$$ and the other numbers are all multiples of $$1089$$). See this page for more about the curious number $$1089$$.