AMC12 2025 A

The numerator can be written as $2^{2a} +2^{2a+2k}+...+2^{2a+2mk}$, which is actually a sum of geometric series. This can be expressed as $2^{2a} \cdot \frac{1-2^{2k(m+1)}}{1-2^{2k}}$. The denominator in the same way can be expressed as $2^a \cdot \frac{1-2^{k(m+1)}}{1-2^k}$. Doing some algebra on the top and bottom we get: 2a⋅1+2k(m+1)1+2k=964 The prime factorization of $964$ is $241 \cdot 2^2$. Equating $2^a = 2^2$, we get $a = 2$. Next since $241$ is a prime number we equate the latter half of the product to $241$. 1+2k(m+1)1+2k=241 Doing some algebra we get, that $2^k(2^{km}-241) = 240$ The closest power of $2$ to $241$ is $256$ which is $2^8$. So setting $km = 8$, we get $2^8-241 = 15$ $2^k(15) = 240$, $2^k = 16$, $k = 4$. This we know that if $k = 4$, then $4m = 8$, so $m = 2$. Finally, we conclude that $a = 2$, $m = 2$, and $k = 4$. $4+2+2 = 8$, so the answer is $\boxed{A}$