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 \(1089\).
For reversible factors with more digits, the same sort of reasoning applies. Factors exist only for \(1 \dots 9 \times 9 = 9 \dots 1\) and \(2 \dots 8 \times 4 = 8 \dots 2\). A little experimentation shows that for five-digit numbers there are only \(10,989 \times 9 = 98,901\) and \(21,978 \times 4 = 87,912\). In general for \(n \gt 4\) digits, two reversible factors exist following the pattern \(109 \dots 989\) and twice that, namely, \(219 \dots 978\), where \(9 \dots 9\) represents a series of \((n - 4)\) \(9\)s. In all cases the resulting factors and their corresponding multiples are divisble by \(99\). Likewise, each factor and its multiples is divisible by a number represented by a series of \((n - 2)\) \(1\)s. Whenever \(n\) is even, the resulting factors and multiples are also divisible by \(1089\), the smallest reversible factor.