When is \(\displaystyle{\frac{{{X}}}{{{R}}}}{\left({X}\right)}\) an integer where

Step 1
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 ${\displaystyle \frac{y}{x}=r}$ has only one digit.
Trying ${\displaystyle r=2}$, we observe that in the units place we must have ${\displaystyle 2a\equiv bb\text{mod}10}$, and the leading digit must satisfy one of the following:
${\displaystyle 2b=a}$
${\displaystyle 2b+1=a}$
Pairing ${\displaystyle 2a\equiv bb\text{mod}10}$ with ${\displaystyle 2b=a}$ implies ${\displaystyle 4a\equiv ab\text{mod}10}$ forcing ${\displaystyle a=0}$, which is no good (the product should not have a leading digit of 0). Pairing ${\displaystyle 2a\equiv bb\text{mod}10}$ with ${\displaystyle 2b=a}$ gives ${\displaystyle 4a+1\equiv ab\text{mod}10}$ from which we can try ${\displaystyle a=3}$, but then ${\displaystyle 2\times 6+1=13\ne 3}$ (we must have exactly 3, not just 3 mod 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 $\text{Undefined control sequence cancel}$ as these fail to be units mod 10. For ${\displaystyle r=7}$ we can try ${\displaystyle a=3}$, 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.as these fail to be units mod 10. For ${\displaystyle r=7}$ we can try ${\displaystyle a=3}$, 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
${\displaystyle 1089\times 9=9801}$
${\displaystyle 2178\times 4=8712}$
In each case we may split the first two digits from the last two and insert an arbitrary number of 9's, as in
${\displaystyle 1099989\times 9=9899901}$
${\displaystyle 2199978\times 4=8799912}$
with three 9's inserted (these being the only solutions, by the way, with seven digits). There are also solutions like the following:
${\displaystyle 1099109999899989\times 9=9899989999019901}$
${\displaystyle 2199219999789978\times 4=8799879999129912}$
Step 2
Solutions do exist with ratios of 2 or 3 in other bases. Among them:
${\displaystyle 1-12\times 2=2101}$ base three
${\displaystyle 1023\times 3=3201}$ base four
${\displaystyle 13\times 2=31}$ base five
${\displaystyle 4378\times 2=8734}$ base twelve
${\displaystyle 3289\times 3=9823}$ base twelve