2022 AMC 10A Problem 11

Since $4096 = 2^{12}$, we have $\frac{1}{4096} = 2^{-12}$. Convert each radical to a power of $2$:
$$2^m \cdot \sqrt{2^{-12}} = 2^m \cdot 2^{-6} = 2^{m-6}, \qquad 2 \cdot \sqrt[m]{2^{-12}} = 2 \cdot 2^{-12/m} = 2^{1 - 12/m}.$$
Powers of $2$ are equal exactly when their exponents are equal:
$$m - 6 = 1 - \frac{12}{m} \quad\Longrightarrow\quad m^2 - 7m + 12 = 0 \quad\Longrightarrow\quad (m-3)(m-4) = 0.$$
Both $m = 3$ and $m = 4$ are legitimate (neither is $0$), and by Vieta their sum is $7$. Check $m = 3$: $8 \cdot \frac{1}{64} = \frac18$ and $2 \cdot \frac{1}{16} = \frac18$.

The answer is $\boxed{\textbf{(C)}\ 7}$.

Whenever every number in sight is a power of the same base, rewrite in that base and compare exponents; the radical structure disappears and only ordinary algebra remains. The resulting equation in $m$ has $m$ in a denominator, so clear it to get a quadratic, and use Vieta for the sum.

• Mishandling the m-th root, writing (1/4096)^(1/m) as 2^(-12m) or 2^(-m/12) instead of 2^(-12/m).
• Dividing the equation by $m$ carelessly or finding only one root and reporting $3$ or $4$ instead of their sum.