Find $I$ in $\frac{\overline{SIX}}{\overline{NINE}}=\frac23$

Question

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 $I$?

Answer 1

$$3\times\overline{SIX}=2\times\overline{NINE}$$

As you wrote, we have $N=1$. Also, we know that $X$ is even.

Let us separate it into cases. In the following, let $[n]$ be the right-most digit of $n$.

Case 1 : $X=0$. Then, $E=0,5$ from $[2E]=[3X]=0$. If $E=0$, then since the right-most two digits of $3\times\overline{IX}$ has to be $20$, we have to have $[3I]=2$ so $I=4$ which is sufficient. If $E=5$, then $[3I]=3$ so $I=1$ which is not sufficient since $1115$ is not a multiple of $3$.

Case 2 : $X=2$. Then, $E=3,8$. If $E=3$, then $[3I]=2$ so $I=4$. If $E=8$, then $[3I]=3$ so $I=1$ which is not sufficient.

Case 3 : $X=4$. Then, $E=1,6$. If $E=1$, then $[3I]=1$ so $I=7$ which is not sufficient. If $E=6$, then $[3I]=2$ so $I=4$.

Case 4 : $X=6$. Then, $E=4,9$. If $E=4$, then $[3I]=1$ so $I=7$ which is not sufficient. If $E=9$, then $[3I]=2$ so $I=4$.

Case 5 : $X=8$. Then, $E=2,7$. If $E=2$, then $[3I]=0$ so $I=0$ which is not sufficient. If $E=7$, then $[3I]=1$ so $I=7$ which is not sufficient.

Therefore, $\color{red}{I=4}$. $\left(\frac{940}{1410},\frac{942}{1413},\frac{944}{1416},\frac{946}{1419}\right)$

P.S. "every letter denotes a UNIQUE digit" then $\frac{942}{1413}$ is the only solution.

Answer 2

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: $(0,0), (2,3), (4,6), (6,9)$.

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