2025 AMC 10A Problem 5

Question

Consider the sequence of positive integers

\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

2025

th term in this sequence?

5

15

16

44

45

The sequence rises from

1

up to a peak and falls back; group it into blocks by peak value

The block that peaks at

k

has

2k-1

terms and ends at cumulative position

k^2

Group the sequence into blocks. Block

k

reads

k, k-1, \ldots, 2, 1, 2, \ldots, k-1, k,

which is

2k - 1

terms and ends on

k.

So after block

k

we’ve used

1 + 3 + \cdots + (2k-1) = k^2

terms. Notice

2025 = 45^2.

That’s exactly the end of block

45,

whose last term is

45.

Thus, E is the correct answer.

Problem 5 in other years