TITLE: Find $I$ in $\frac{\overline{SIX}}{\overline{NINE}}=\frac23$
QUESTION [5 upvotes]: In $\frac{\overline{SIX}}{\overline{NINE}}=\frac23$ every letter denotes a  UNIQUE digit,find $I$.    

Expanding the fraction in base $10$ we have:
$300S+30I+3X=2020N+200I+2E$ , but this doesn't help much.We may deduce that since $\overline{SIX}\lt1000$ so $\overline{NINE}\lt1500$ , now we know $N=1$.How to find other digits specially $I$?

REPLY [2 votes]: Because $N=1$, the equation is:
$300S + 3X = 2020 + 170I +2E$
Let's write it $R(S,X) = L(I,E)$.
We notice that $L(5,0)=2870$ and $R(9,9)=2727 < L(5,0)$ so $I<5$.
Let's check the remaining 5 possible values for $I$:

$I=0$    

Then $2020 \leq L(0,E) \leq 2038$
But $R(7,0)=2100$ and $R(6,9) = 1827$ so we cannot have $R(S,X) = L(0,E)$.
Hence $I=0$ is not possible.

$I=1$   

Then $2190 \leq L(1,E) \leq 2208$
But $R(8,0)=2400$ and $R(7,9) = 2127$ so we cannot have $R(S,X) = L(1,E)$.
Hence $I=1$ is not possible.

$I=2$   

Then $2360\leq L(2,E) \leq 2378$
But $R(8,0)=2400$ and $R(7,9) = 2127$ so we cannot have $R(S,X) = L(2,E)$.
Hence $I=2$ is not possible.

$I=3$   

Then $2530\leq L(3,E) \leq 2548$
But $R(9,0)=2700$ and $R(8,9) = 2427$ so we cannot have $R(S,X) = L(3,E)$.
Hence $I=3$ is not possible.

$I=4$   

Then $2700\leq L(4,E) \leq 2718$
We have $R(8,9) = 2427 < 2700$ so we must have $S=9$.
The equation becomes $3X=2E$ . 
 We find 4 solutions: $(0,0), (2,3), (4,6), (6,9)$.

Finally, $I=4$ and the 4 solutions are:
$\left(\frac{940}{1410},\frac{942}{1413},\frac{944}{1416},\frac{946}{1419}\right)$.