2025 AMC 12A Problem 21

If $k=0$ or $m=0,$ the original ratio is just $2^a,$ which cannot equal $964.$ Hence $k,m\gt0.$ Summing the geometric series, the numerator is $4^a\dfrac{4^{k(m+1)} - 1}{4^k - 1}$ and the denominator is $2^a\dfrac{2^{k(m+1)} - 1}{2^k - 1}.$ Using $4^N - 1 = (2^N - 1)(2^N + 1),$ the ratio simplifies to $$2^a \cdot \frac{2^{k(m+1)} + 1}{2^k + 1} = 964.$$ The fraction on the left is odd over odd, so its power of $2$ is $0.$ Since $964=2^2\cdot241,$ we must have $a=2,$ and $$2^{k(m+1)}+1=241(2^k+1).$$ Reducing this equation modulo $2^k$ shows that $240$ is divisible by $2^k,$ so $k\le4.$ For $k=1,2,3,$ the required powers of $2$ would be $722,1204,2168,$ none of which is a power of $2.$ For $k=4,$ however, $241(2^4+1)=4097=2^{12}+1,$ so $k(m+1)=12$ and $m=2.$ This also proves uniqueness. Then $a + k + m = 2 + 4 + 2 = 8.$ Thus, the correct answer is A.