2025 AMC 10A Problem 5

Cut the sequence after every $1$:
$$1 \mid 2,1 \mid 2,3,2,1 \mid 2,3,4,3,2,1 \mid 2,3,4,5,4,3,2,1 \mid \dots$$
Block $n$ climbs $2, 3, \dots, n$ and then descends $n-1, \dots, 1$, so it has $(n-1) + (n-1) = 2n - 2$ terms for $n \ge 2$, while block $1$ is the single term $1$.

The number of terms in the first $n$ blocks is therefore
$$1 + \sum_{k=2}^{n} (2k-2) = 1 + 2\cdot\frac{(n-1)n}{2} = n(n-1) + 1.$$
Check: $n = 4$ gives $13$, and the sequence does read $\dots,2,1$ at positions $12, 13$.

Now locate $2025$. For $n = 45$ the running total is $45 \cdot 44 + 1 = 1981$, and for $n = 46$ it is $46 \cdot 45 + 1 = 2071$. So term $2025$ lies in block $46$, at position
$$2025 - 1981 = 44$$
inside that block. Block $46$ begins $2, 3, 4, \dots$, climbing to its peak $46$ at its $45$th entry, so its $j$th entry is $j + 1$ for $j \le 45$. With $j = 44$ the value is $45$.

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

Any sequence built from blocks of growing length yields to the same two steps: put the block lengths in closed form, then find the last block boundary before the target and count forward. Writing the cumulative total as $n(n-1)+1$ turns the search for the right block into a single inequality rather than a long addition, and it is worth checking that formula against a small case before trusting it at $n = 46$.

Test the perfect-square positions on the visible part of the sequence: term $4$ is $2$, term $9$ is $3$, term $16$ is $4$, term $25$ is $5$. The term in position $m^2$ is $m$. Since $2025 = 45^2$, the answer is $45$. The reason is the count above: the peak $n$ of block $n$ sits at position $n(n-1) + 1 - (n-1) = (n-1)^2 + 1$, so the entry just before it, namely $n - 1$, sits at position $(n-1)^2$.

• Landing in the right block but reading its 44th entry as 44; the block starts at 2, so its 44th entry is 45.
• Giving block $n$ a length of $2n$ instead of $2n - 2$, which shifts every boundary and every answer.
• Forgetting that block $1$ is a single term, so that the cumulative totals are off by one throughout.