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2022 AMC 10B Problem 25

Problem 25 of 25HarderNumber Theory

Let x0,x_0, x1,x_1, x2,x_2, \dotsc be a sequence of numbers, where each xkx_k is either 00 or 1.1. For each positive integer n,n, define Sn=k=0n1xk2kS_n = \sum_{k=0}^{n-1} x_k 2^k Suppose 7Sn1(mod2n)7S_n \equiv 1 \pmod{2^n} for all n1.n \geq 1. What is the value of the sum x2019+2x2020+x_{2019} + 2x_{2020} + 4x2021+8x2022?4x_{2021} + 8x_{2022}?

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Solution

The desired sum is S2023S201922019.\frac{S_{2023}-S_{2019}}{2^{2019}}. Also, 0Sn<2n.0\le S_n<2^n. Therefore, for a unique mn{0,1,,6},m_n\in\{0,1,\ldots,6\}, 7Sn=mn2n+1.7S_n=m_n2^n+1. Reducing modulo 77 gives mn2n1(mod7).m_n2^n\equiv-1\pmod7. Since 231(mod7)2^3\equiv1\pmod7 and 20190(mod3),2019\equiv0\pmod3, we get m2019=6.m_{2019}=6. Since 20231(mod3),2023\equiv1\pmod3, we have 2m20231(mod7),2m_{2023}\equiv-1\pmod7, so m2023=3.m_{2023}=3. Hence S2019=622019+17,S2023=322023+17.\begin{gathered}S_{2019}=\frac{6\cdot2^{2019}+1}{7},\\ S_{2023}=\frac{3\cdot2^{2023}+1}{7}.\end{gathered} Finally, S2023S201922019=32467=6. \frac{S_{2023}-S_{2019}}{2^{2019}} =\frac{3\cdot2^4-6}{7}=6. Thus, the correct answer is A .

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Concepts: modular arithmetic · number base · power of 2

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.