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2025 AMC 10A Problem 5

Problem 5 of 25EasierNumber Theory

Consider the sequence of positive integers 1,2,1,2,3,2,1,2,3,4,3,2,1,2,3,4,5,4,3,2,1,2,3,4,5,6,5,4,3,2,1,2,… \begin{gathered} 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, \\ 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, \\ 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, \\ 1, 2, \ldots \end{gathered} What is the 20252025th term in this sequence?

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Solution

Group the sequence into blocks. Block kk reads k,k−1,…,2,1,2,…,k−1,k,k, k-1, \ldots, 2, 1, 2, \ldots, k-1, k, which is 2k−12k - 1 terms and ends on k.k. So after block kk we’ve used 1+3+⋯+(2k−1)=k21 + 3 + \cdots + (2k-1) = k^2 terms. Notice 2025=452.2025 = 45^2. That’s exactly the end of block 45,45, whose last term is 45.45. Thus, E is the correct answer.

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Concepts: pattern recognition · perfect square

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