AMC 12A 2025 (Problem 21)There is a unique ordered triple (a,k,m)(a,k,m)(a,k,m) of nonnegative integers such that 4a+4a+k+4a+2k+⋯+4a+mk2a+2a+k+2a+2k+⋯+2a+mk=964\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}}=9642a+2a+k+2a+2k+⋯+2a+mk4a+4a+k+4a+2k+⋯+4a+mk=964. What is a+k+ma+k+ma+k+m?(A) 8\text{(A)}\;8(A)8(B) 9\text{(B)}\;9(B)9(C) 10\text{(C)}\;10(C)10(D) 11\text{(D)}\;11(D)11(E) 12\text{(E)}\;12(E)12Related TopicsCoreToolkit 4 — Geometric Sequence and SeriesMinorToolkit 10 — Difference of squaresHints (6)Hint 14a+4a+k+4a+2k+⋯+4a+mk2a+2a+k+2a+2k+⋯+2a+mk=964\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}}=9642a+2a+k+2a+2k+⋯+2a+mk4a+4a+k+4a+2k+⋯+4a+mk=9644a+4a+k+4a+2k+⋯+4a+mk=964(2a+2a+k+2a+2k+⋯+2a+mk)4^a+4^{a+k}+4^{a+2k}+\cdots+4^{a+mk}=964\left(2^a+2^{a+k}+2^{a+2k}+\cdots+2^{a+mk}\right)4a+4a+k+4a+2k+⋯+4a+mk=964(2a+2a+k+2a+2k+⋯+2a+mk)4a(1+4k+(4k)2+⋯+(4k)m)=964⋅2a(1+2k+(2k)2+⋯+(2k)m)4^a\left(1+4^k+(4^k)^2+\cdots+(4^k)^m\right)=964\cdot2^a\left(1+2^k+(2^k)^2+\cdots+(2^k)^m\right)4a(1+4k+(4k)2+⋯+(4k)m)=964⋅2a(1+2k+(2k)2+⋯+(2k)m)By Toolkit 4 — Geometric Sequence and Series:2a⋅(4k)m+1−14k−1=964((2k)m+1−12k−1)2^a\cdot\frac{(4^k)^{m+1}-1}{4^k-1}=964\left(\frac{(2^k)^{m+1}-1}{2^k-1}\right)2a⋅4k−1(4k)m+1−1=964(2k−1(2k)m+1−1)Hint 22a⋅(2k(m+1))2−1(2k)2−1=964(2k(m+1)−12k−1)2^a\cdot\frac{\left(2^{k(m+1)}\right)^2-1}{(2^k)^2-1}=964\left(\frac{2^{k(m+1)}-1}{2^k-1}\right)2a⋅(2k)2−1(2k(m+1))2−1=964(2k−12k(m+1)−1)By Toolkit 10 — Difference of squares:2a⋅2k(m+1)+12k+1=9642^a\cdot\frac{2^{k(m+1)}+1}{2^k+1}=9642a⋅2k+12k(m+1)+1=964⟹2a(2k(m+1)+1)=964(2k+1)\Longrightarrow 2^a\left(2^{k(m+1)}+1\right)=964(2^k+1)⟹2a(2k(m+1)+1)=964(2k+1)Hint 3If k=0k=0k=0,2a(1+1)=964(2)2^a(1+1)=964(2)2a(1+1)=964(2)which is impossible.So k≥1k\ge1k≥1.Hint 42a((2k)m+1+1)=4×241(2k+1)2^a\left((2^k)^{m+1}+1\right)=4\times241(2^k+1)2a((2k)m+1+1)=4×241(2k+1)Even parts:2a=4⟹a=2(1)2^a=4\Longrightarrow a=2\qquad(1)2a=4⟹a=2(1)Odd parts:(2k)m+1+1=241(2k+1)(2^k)^{m+1}+1=241(2^k+1)(2k)m+1+1=241(2k+1)Hint 52mk+k+1=241⋅2k+2412^{mk+k}+1=241\cdot2^k+2412mk+k+1=241⋅2k+241⟹2k(2mk−241)=240=16×15\Longrightarrow 2^k(2^{mk}-241)=240=16\times15⟹2k(2mk−241)=240=16×15⟹2k=16⟹k=4(2)\Longrightarrow2^k=16\Longrightarrow k=4\qquad(2)⟹2k=16⟹k=4(2)2mk−241=15⟹24m=256=282^{mk}-241=15\Longrightarrow2^{4m}=256=2^82mk−241=15⟹24m=256=28⟹4m=8⟹m=2(3)\Longrightarrow4m=8\Longrightarrow m=2\qquad(3)⟹4m=8⟹m=2(3)Hint 6From (1)(1)(1), (2)(2)(2), and (3)(3)(3):a+k+m=2+4+2=8a+k+m=2+4+2=8a+k+m=2+4+2=8Final Answer(A) 888Related Problems (3)AIME II 2026 (Problem 9)AMC 10A 2025 (Problem 11)AMC 10A/12A 2025 (Problem 13/5)