Problem:
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,…
What is the 2025th term in this sequence?
Answer Choices:
A. 5
B. 15
C. 16
D. 44
E. 45
Solution:
We observe the sequence forms blocks as follows:
1234→1,2,→1,2,3,2,→1,2,3,4,3,2,→1,2,3,4,5,4,3,2, and so on.
Each nth block contains 2n terms.
The total number of terms up to the nth block is
2(1+2+3+⋯+n)=2×2n(n+1)=n(n+1).
We require that
n(n+1)≤2025.
This gives
n2+n−2025≤0.
Approximating n, we get
n≈44since44×45=1980<2025<45×46=2070.
Hence, the 44th sequence has 88 terms and the 45th sequence has 90 terms.
The 2025th term lies in the 45th sequence, and it is the
2025−1980=45th term of that sequence.
In the 45th sequence
1,2,3,…,46,45,…,2,
the 45th term is clearly
(E) 45
The problems on this page are the property of the MAA's American Mathematics Competitions