2019 AMC 10A Problem 18

Step 1: remember the base-ten version. In base ten, $0.\overline{23} = \frac{23}{99}$. Here is why: if $x = 0.2323\ldots$, then $100x = 23.2323\ldots$. Subtracting, $99x = 23$.

Step 2: do the same in base $k$.

1. In base $k$, moving the digits two places to the left means multiplying by $k^2$ (just as it means multiplying by $10^2 = 100$ in base ten).
2. Let $x = 0.\overline{23}_k$. Then $k^2 x = 23.\overline{23}_k$.
3. The number written $23_k$ means $2 \cdot k + 3$.
4. Subtract: $k^2 x - x = 2k + 3$, so
$$x = \frac{2k + 3}{k^2 - 1}.$$

Step 3: match it to $\frac{7}{51}$.

1. We need $\dfrac{2k + 3}{k^2 - 1} = \dfrac{7}{51}$, which means $51(2k + 3) = 7(k^2 - 1)$.
2. The left side is a multiple of $51$, so the right side $7(k^2 - 1)$ must be too. Since $7$ shares no factor with $51 = 3 \cdot 17$, the $51$ has to go into $k^2 - 1$. So $k^2 - 1$ is a multiple of $51$.
3. Test the choices:

Only $255 = 5 \cdot 51$ is a multiple of $51$, so $k = 16$.
8. Check. $2 \cdot 16 + 3 = 35$, and $\dfrac{35}{255} = \dfrac{7 \cdot 5}{51 \cdot 5} = \dfrac{7}{51}$. It works.

The answer is $\boxed{\textbf{(D)}\ 16}$.

The "multiply and subtract" trick for repeating decimals works in any base; only the multiplier changes from $100$ to $k^2$. Once the equation is written, a divisibility clue ($51$ must divide $k^2 - 1$) picks out the answer from the choices faster than solving the equation.

Solve the equation directly. $51(2k + 3) = 7(k^2 - 1)$ becomes $102k + 153 = 7k^2 - 7$, or
$$7k^2 - 102k - 160 = 0.$$
This factors as $(7k + 10)(k - 16) = 0$. Since $k$ is positive, $k = 16$.