2025 AMC 10A Problem 5

1. The sequence goes up and back down again and again. Cut it right after each $1$ to see the pattern:
$$1 \;\big|\; 2,1 \;\big|\; 2,3,2,1 \;\big|\; 2,3,4,3,2,1 \;\big|\; 2,3,4,5,4,3,2,1 \;\big|\; \dots$$
Call these groups $1, 2, 3, 4, 5, \ldots$ Group $n$ climbs from $2$ up to $n$, then comes back down to $1$.

1. Count the lengths: group $1$ has $1$ term, and then groups $2, 3, 4, 5$ have $2, 4, 6, 8$ terms. Group $n$ has $2n - 2$ terms.

1. Add up to see where each group ends:

These totals are $1 \cdot 0 + 1,\ 2 \cdot 1 + 1,\ 3 \cdot 2 + 1,\ 4 \cdot 3 + 1,\ 5 \cdot 4 + 1$. So the first $n$ groups hold $n(n-1) + 1$ terms. (Reason: $2 + 4 + \cdots + (2n-2)$ is twice $1 + 2 + \cdots + (n-1)$, which is $n(n-1)$.)

1. We want a total near $2025$, so try $n$ near $45$. First $45$ groups: $45 \times 44 + 1 = 1981$ terms. First $46$ groups: $46 \times 45 + 1 = 2071$ terms. So term $2025$ is inside group $46$.

1. How far into group $46$? $2025 - 1981 = 44$, so it is the $44$th number of group $46$.

1. Group $46$ goes $2, 3, 4, \ldots$ up to $46$. Its 1st number is $2$, its 2nd is $3$, so its $j$th number is $j + 1$ (while still climbing). The $44$th number is $45$.

The answer is $\boxed{\textbf{(E)}\ 45}$.

For a sequence built from groups that keep getting longer, find a formula for "how many terms so far", test it on the first few groups, then find the last group that ends before your target and count forward. Checking the formula on small groups (like $13$ terms after group $4$) catches off-by-one errors before they matter.

Look at the square-numbered positions in the sequence given: term $1$ is $1$, term $4$ is $2$, term $9$ is $3$, term $16$ is $4$, term $25$ is $5$. It looks like term number $m \times m$ equals $m$. This holds every time: group $m + 1$ reaches the value $m$ at position $m(m-1) + 1 + (m - 1) = m^2$. Since $2025 = 45 \times 45$, term $2025$ is $45$.