When is $X/R(X)$ an integer where $R(X)$ is the reverse of an integer $X$?

Question

My question concerns reverse numbers (e.g. $1234 → 4321$). Is it possible to find integer solutions greater than $1$ for such numbers when you take their ratio? I am not interested in trivial solutions such as powers of ten and their multiples ($0, 100, 20, 1100$ etc.). Let's say you have a number $X$ and $R(X)$ so that $X/R(X) = n$. I've done some testing and have not found any solutions for $n = 2,3$ up to $X = 10^7$.

I'm fairly certain that I have proved that such a number cannot exist if it has an odd number of digits.

Any ideas or solutions? (This is just an idea that occurred to me, nothing I really have to solve)

Thanks!

Answer 1

Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did a quick search where I look for values of $n$ that satisfy $12\le n\le10^9$

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
ParallelTable[
  If[IntegerQ[n/IntegerReverse[n]]&&IntegerQ[n/10]==False&& 
    PalindromeQ[n]==False,n,Nothing],{n,12,10^9}]//.{}-> 
  Nothing

Running the code gives:

Out[1]={8712, 9801, 87912, 98901, 879912, 989901, 8799912, 9899901,
87128712, 87999912, 98019801, 98999901, 871208712, 879999912,
980109801, 989999901}

Where I removed the palindromic numbers and numbers of the form $10^k$ for $k\in\mathbb{N}$.

Answer 2

There are no solutions with a ratio of 2 or 3. In fact the only possible ratios are indeed 4 and 9.

Let $x$ be the smaller number with terminal digit $a$, and $y$ be thevlarger number either terminal digit $b$. Then $y$ has more digits than $x$, which kills us, unless the ratio $y/x=r$ has only one digit.

Trying $r=2$, we observe that in the units place we must have $2a\equiv b\bmod 10$, and the leading digit must satisfy one of the following:

$2b=a$

$2b+1=a$

Pairing $2a\equiv b\bmod 10$ with $2b=a$ implies $4a\equiv a\bmod 10$ forcing $a=0$, which is no good (the product should not have a leading digit of $0$). Pairing$ 2a\equiv b\bmod 10$ with $2b=a$ gives $4a+1\equiv a\bmod 10$ from which we can try $a=3$, but then $2×6+1=13\not=3$ (we must have exactly $3$, not just $3\bmod 10$, or else $y$ has too many digits).

Similar logic eliminates $3$ as a possible ratio.

For ratios from $5$ to $8$, we use the fact that $b$, which must be the leading digit of $x$ as well as the last digit of $y$, can only be $1$. Thus $r\not\in\{5,6,8\}$ as these fail to be units $\bmod 10$. For $r=7$ we can try $a=3$, but then the leading digit of $y$ would have to be $3$ whereas the actual product $7x$, if it has the same digits as $x$, has to begin with $7,8,$ or $9$.

So we are left with the ratios that are well known to have solutions, $4$ and $9$. In both cases there are infinitely many solutions. The minimal ones for each ratio are of course

$1089×9=9801$

$2178×4=8712$

In each case we may split the first two digits from the last two and insert an arbitrary number of $9$'s, as in

$1099989×9=9899901$

$2199978×4=8799912$

with three $9$'s inserted (these being the only solutions, by the way, with seven digits). There are also solutions like the following:

$1099109999899989×9=9899989999019901$

$2199219999789978×4=8799879999129912$

---

Solutions do exist with ratios of $2$ or $3$ in other bases. Among them:

$1012×2=2101\text{ base three}$

$1023×3=3201\text{ base four}$

$13×2=31\text{ base five}$

$4378×2=8734\text{ base twelve}$

$3289×3=9823\text{ base twelve}$

Answer 3

Suppose we have $n*abcd=dcba$.

Just looking at the leftmost digit, we have $d=na+m$ where $m$ is the amount carried over from the digit to the right. Note that $m<n$.

Looking at the rightmost digit, we have $10p+a=nd$ where $p$ is the amount carried over to the left. Again we have $p<n$.

Combining these two equations you get $10p+a=n(na+m)$ which simplifies to $(n^2-1)a = 10p-nm$.

Case n=2:
The equation becomes $3a = 10p-2m$ with $p,m\in\{0,1\}$. This is easily verified to have no integer solutions.

Case n=3:
The equation becomes $8a = 10p-3m$, with $p,m\in\{0,1,2\}$. Again this is easily checked to have no solutions with $a$ being a single digit.

Note that this argument works for any length of numbers, so there are no solutions with $n=2$ or $n=3$.