2021 AMC 10B Problem 8

Read the pattern off the diagram: $1$, $9$ and $25$ each sit at the top-right corner of the $1\times1$, $3\times3$ and $5\times5$ block centered at $1$. So every odd square $(2k+1)^2$ finishes its layer at the top-right corner of a $(2k+1)\times(2k+1)$ block, and the next number starts the new layer one cell to the right, then runs down, then left along the bottom, then up, then right along the top.

The $13 \times 13$ block occupies rows $2$ through $14$ and columns $2$ through $14$, so $169$ is in row $2$, column $14$. Its top row (row $2$, columns $2$–$14$) reads $157, 158, \dots, 169$.

The outer layer: $170$ goes to row $2$, column $15$; $171$–$183$ run down the right column; $184$–$197$ run left along row $15$; $198$–$211$ run up column $1$, ending with $211$ in the top-left corner; and $212$–$225$ fill row $1$. Thus row $2$, column $1$ holds $210$.

Row $2$ from left to right: $210, 157, 158, \dots, 169, 170$. Its least number is $157$ and its greatest is $210$, with sum $367$.

The answer is $\boxed{\textbf{(A)}\ 367}$.

In a square spiral, the squares of consecutive odd numbers mark the corners where layers close, and everything else is offset from those anchors. Locate the two layers that touch the target row (the $13\times13$ block's top edge and the $15\times15$ block's sides), then count from the nearest anchor. Verify the direction of travel from the given diagram rather than from memory.

• Assuming the second row is 15 consecutive numbers, or misplacing the corners and reading off 211 + 157 = 368 or 212 + 157 = 369.
• Putting odd squares in the wrong corner (e.g. bottom-right), which shifts every number in the row.