2025 AMC 10A Problems/Problem 5

Consider the sequence of positive integers $1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2 \dots$

What is the $2025$ th term in the sequence?

$\textbf{(A) } 5 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 16 \qquad\textbf{(D) } 44 \qquad\textbf{(E) } 45$

Solution 1

One possible way the sequence could've been constructed was by putting "mountains" going up from $1$ , to $n+1,$ then going back down to $2.$ For example, the first few "mountains" look like this:

$12|1232|123432|12345432|...$

So, the $n^{th}$ mountain has length $2n$ and has highest number $n+1.$ We want to add mountains until we get a total length as close as possible, but not exceeding, $2025.$ Let the last mountain we sum be mountain $a.$ Hence, $2+4+6+...+2a=2(1+2+3+...+a)=a(a+1)\le2025$ $\implies a^2<2025\implies a<45,$ so our max $a$ is $44.$ In this $44^{th}$ mountain, the max number is $45,$ so the $45^{th}$ mountain has max number $46.$ Next, $44(44+1)=1980,$ so we're looking for the $45^{th}$ number in the $45^{th}$ mountain, which is $\boxed{\text{(E) }45}.$

~Tacos_are_yummy_1

Page

Toolbox

Search

2025 AMC 10A Problems/Problem 5

Solution 1