2025 AMC 12A Problems/Problem 21

Problem 21

There is a unique ordered triple $(a,k,m)$ of nonnegative integers such that $\frac{4^a + 4^{a+k}+4^{a+2k}+\cdots + 4^{a+mk}}{2^a + 2^{a+k} + 2^{a+2k}+ \cdots + 2^{a+mk}} = 964.$ What is $a+k+m$ ?

$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C)} 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$

Solution 1

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:

\begin{equation} 2^a \cdot \frac{1+2^{k(m+1)}}{1+2^k} = 964 \end{equation}

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 $241$ . \begin{equation} \frac{1+2^{k(m+1)}}{1+2^k} = 241 \end{equation} 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 $\fbox{A}$

~MATHEMATICIAN635 ~Minor Edits by MALICIOUSFISH23

Actually when it comes to $2^k(2^{km}-241) = 240$ ; we can directly observe the answer by discovering $2^k$ is even and $2^{km}-241$ must be odd. So, the only available divisors of 240 is 16 and 15. And we can have $m = 2$ , and $k = 4$ .

~DRA777

Video Solution 1 by OmegaLearn

https://youtu.be/Cx8BScqcx6U

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=Y9sG6vZPDQo

2025 AMC 12A (Problems • Answer Key • Resources)
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2025 AMC 12A Problems/Problem 21

Contents

Problem 21

Solution 1

Video Solution 1 by OmegaLearn

Video Solution 2 by SpreadTheMathLove

See Also