2019 AMC 10A Problem 18

Let $x = 0.\overline{23}_k$. Multiplying by $k^2$ shifts the repeating block two places: $k^2 x = 23_k + x$, where $23_k = 2k + 3$. So
$$x = \frac{2k + 3}{k^2 - 1}.$$
We need $\frac{2k+3}{k^2-1} = \frac{7}{51}$, i.e. $51(2k + 3) = 7(k^2 - 1)$.

Fast route: the right side is a multiple of $7$ and $\gcd(7, 51) = 1$, so $51 \mid k^2 - 1$. Among the choices, $k^2 - 1$ is $168, 195, 224, 255, 288$, and only $255 = 5 \cdot 51$ is a multiple of $51$. So $k = 16$; check $2k + 3 = 35 = 7 \cdot 5$, and indeed $\frac{35}{255} = \frac{7}{51}$.

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

The base-ten fact $0.\overline{ab} = \frac{10a + b}{99}$ is really $\frac{\text{block}}{k^2 - 1}$ in any base $k$, derived by the same shift-and-subtract trick. Once the equation is set up, the divisibility structure ($51$ must divide $k^2 - 1$) is quicker than expanding a quadratic; the answer choices exist to be tested this way.

Solve directly: $7k^2 - 102k - 160 = 0$. The discriminant is $102^2 + 4 \cdot 7 \cdot 160 = 10404 + 4480 = 14884 = 122^2$, so $k = \frac{102 + 122}{14} = 16$. Alternatively factor as $(7k + 10)(k - 16) = 0$.

• Writing the repeating value as (2k + 3)/k^2 or 23/99 style with the wrong denominator, or solving the quadratic and mis-simplifying.
• Reading $23_k$ as the number $23$ instead of $2k + 3$.