March Break 2024 - Problem 7


Let an=(n1)!+n22n+1a_n = (n - 1)! + n^2 - 2n + 1 for n1n \geq 1. Let bnb_n be defined as the remainder when ana_n is divided by nn. Compute n=17b2n1\sum\limits_{n = 1}^{7} b_{2n - 1}.

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Problem Tags: Algebra Number theory

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Category: March Break Contest
Points: 5
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