Take any n-digit integer (n > 1) and interchange its first and last digits. If neither of these is 0, and they are different, does it happen infinitely often that the resulting number is a multiple of the original one?
If so, for each n (say up to n=20, unless a general solution is provided), what is the largest and smallest numbers for which this happens?
Suppose the first digit is a and the last digit is c. (The rest of the number can be denoted as b.)
Then our original number is $a\cdot 10^{n-1}+b\cdot 10 + c$ and after swapping the digits it is $c\cdot 10^{n-1}+b\cdot 10 + a$.
The question asks whether there are $a,b,c$ such that $\frac{c\cdot 10^{n-1}+b\cdot 10 + a}{a\cdot 10^{n-1}+b\cdot 10 + c} = N$ for some integer $N > 1$. Clearly $N < 10$ as they have the same number of digits as well. Also, $c > a$.
Subtracting one from the above equation, we get $\frac{(10^{n-1}-1)(c-a)}{a\cdot 10^{n-1}+b\cdot 10 + c} = N-1$, i.e. $\frac{(10^{n-1}-1)(c-a)}{N-1} = a\cdot 10^{n-1}+b\cdot 10 + c$.
Now we proceed via casework:
We have the bounds that $N-1 \in \{1,2,3,4,5,6,7,8\}$.
If $N-1 \in \{1,2,4,5,8\}$, then $10^{n-1}-1$ shares no factors with $N-1$ so if the LHS is an integer, it must be a factor of $10^{n-1}-1$. Since $a$ should be smaller than $c$, our options are: $19\dots98$, $29\dots97$, $39\dots96$, $49\dots95$, corresponding to $\frac{c-a}{N-1}$ values of $2,3,4,5$ but at the same time $c-a$ values of $7,5,3,1$. At no point are the right numbers a multiple of the left numbers, so this rules out this possibility.
If $N-1 \in \{3,6\}$, then $10^{n-1}-1$ could absorb a factor of $3$. This gives our options as $13\dots32$, $16\dots65$, $19\dots98$, $26\dots64$, $29\dots97$, corresponding to $\frac{3(c-a)}{N-1}$ values of $4,5,6,8,9$ and $c-a$ values of $1,4,7,2,5$. But similarly, the right hand numbers are never a multiple of the left hand numbers.
If $N-1 = 7$, then $10^{n-1}-1$ could absorb a factor of $7$. We're pretty restricted in this case - $(c-a)$ must be $8$, because otherwise the LHS has $n-1$ digits versus the RHS $n$ digits. $7|10^{n-1}-1$ only when $6|n-1$ for which the result is $142857\dots142857$. Adding $9\dots9$ to this, we get $1142857\dots142856$, which has $c-a$ value of $5$, which is not $8$. So this concludes our casework and therefore there are no solutions.
I don't think there are any solutions
To see this. Let's try to figure out what possible first and last digits are. Let those digits be $x$ and $y$ where $x > y$. That means the numbers would be $xd_0d_1\ldots y$ (this is concatenation not multiplication) and $yd_0d_1\ldots x$ where $(yd_0d_1\ldots x) * k = (xd_0d_1\ldots y)$.
We can draw a few useful facts from this. $\left \lfloor{x\div k}\right \rfloor= y$ (looking at the most significant digit) and $x*k = y\text{ }(\text{mod }10)$ (looking at the least significant digit). So, let's see what combinations qualify. For the table below the numbers at the top are different values for $x$ and the left numbers are $k$. Solving the two equations for $y$ give the left and right numbers in each cell.| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | --+-----------------------------------------------+ 2 | 1 4 | 1 6 | 2 8 | 2 0 | 3 2 | 3 4 | 4 6 | 4 8 | 3 | 0 6 | 1 9 | 1 2 | 1 5 | 2 8 | 2 1 | 2 4 | 3 7 | 4 | 0 8 | 0 2 | 1 6 | 1 0 | 1 4 | 1 8 | 2 2 | 2 6 | 5 | 0 0 | 0 5 | 0 0 | 1 5 | 1 0 | 1 5 | 1 0 | 1 5 | 6 | 0 2 | 0 8 | 0 4 | 0 0 | 1 6 | 1 2 | 1 8 | 1 4 | 7 | 0 4 | 0 1 | 0 8 | 0 5 | 0 2 | 1 9 | 1 6 | 1 3 | 8 | 0 6 | 0 4 | 0 2 | 0 0 | 0 8 | 0 6 | 1 4 | 1 2 | 9 | 0 8 | 0 7 | 0 6 | 0 5 | 0 4 | 0 3 | 0 2 | 1 1 |
Continuing...
Looking at the table above we know that the valid combinations must result in the same number (also it can't be 0 because 0 is not valid for $y$). This leaves us two options. $x=8, y=2, k=4$ or $x=9, y=1, k=9$.
However, in both of these cases $x$ is divisible by $k$ which is a problem. Looking back at the equation $(yd_0d_1\ldots x) * k = (xd_0d_1\ldots y)$, we know that $(d_0d_1\ldots x) * k = (d_0d_1\ldots y)$ as long as $y*k = x$. But we know that $(d_0d_1\ldots x) > (d_0d_1\ldots y)$ which means there are no solutions for $(d_0d_1\ldots x) * k = (d_0d_1\ldots y)$ and therefore no solutions to this puzzle.
