Art of Problem Solving

2025 AMC 12A Problems/Problem 21: Difference between revisions

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Finally, we conclude that <imath>a = 2</imath>, <imath>m = 2</imath>, and <imath>k = 4</imath>.
Finally, we conclude that <imath>a = 2</imath>, <imath>m = 2</imath>, and <imath>k = 4</imath>.


4+2+2 = 8, so the answer is A
4+2+2 = 8, so the answer is <imath>\fbox{A}</imath>


~MATHEMATICIAN635
~MATHEMATICIAN635
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==Video Solution 1 by OmegaLearn==
==Video Solution 1 by OmegaLearn==
https://youtu.be/Cx8BScqcx6U
https://youtu.be/Cx8BScqcx6U
==Video Solution 2 by SpreadTheMathLove==
https://www.youtube.com/watch?v=Y9sG6vZPDQo


==See Also==
==See Also==

Latest revision as of 01:21, 11 November 2025

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

See Also

2025 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.