2025 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 5 Easy

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,\cdots

What is the 2025\text{th} term in this sequence?

  • A.

    5

  • B.

    15

  • C.

    16

  • D.

    44

  • E.

    45

Answer:E

Divide the whole sequence at each 1.

The n-th sequence would be: 1, 2, \dots, n+1, \dots, 2, which has length 2n.

The cumulative length until the end of the n-th sequence is: f(n) = \sum_{k=1}^{n} 2k = 2 \cdot \frac{n(n+1)}{2} = n(n+1)

Computing: f(44) = 44 \times 45 = 1980 and f(45) = 45 \times 46 = 2070.

Hence, the 2025th term appears at position (2025 - 1980) = 45 within the 45th sequence 1, 2, \ldots, 46, \ldots, 2, giving us 45.

Related Problems
Problem 3 2025 AMC 10 A
Problem 4 2025 AMC 10 A
Problem 6 2025 AMC 10 A
Problem 7 2025 AMC 10 A
Think Academy Powered by ChatGPT