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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?

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    $\begingroup$ I could not find a single occurrence to think on. Do you have any example? :P $\endgroup$
    – Hasan
    Jun 19 '20 at 16:50
  • $\begingroup$ I hope this doesn't count as a clue, but to save people some work, there are no 2-digit numbers with this property. $\endgroup$
    – Sputnik
    Jun 19 '20 at 18:23
  • $\begingroup$ I don't think I've ever seen a number with the stated property - but I think it's universal that the difference between the number and the result of the described operation will be a multiple of 9 (i.e., ab......cd - db......ca is always 9 × k). $\endgroup$ Jun 19 '20 at 18:42
  • $\begingroup$ My computer thinks there is no such number up to 1000000000. $\endgroup$
    – Florian F
    Jun 19 '20 at 19:26
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    $\begingroup$ Dear commenters: what happened to not putting spoilers in the comments? Also: math.stackexchange.com/questions/2860037/… $\endgroup$
    – Bass
    Jun 19 '20 at 21:49
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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.

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  • $\begingroup$ I like how we used very different methods to come to the same conclusion. $\endgroup$ Jun 19 '20 at 21:52
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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.

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  • $\begingroup$ I couldn't figure out the formatting to make it work in one spoiler $\endgroup$ Jun 19 '20 at 19:59

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