2025 AMC 10A Problem 19

Each interior entry feeds two entries below, and the end values $-1$ and $1$ cancel in the sum. So every row’s total doubles the one above. The top row sums to $3,$ and $12{,}288 = 3 \cdot 2^{12},$ so this is the $12$th row (counting the top as row $0$). Track the diagonals from the left. The second diagonal is $3, 2, 1, 0, -1, \ldots,$ dropping by $1$ each row. The third diagonal adds these up: for $n \ge 4$ it equals $7 - (0 + 1 + \cdots + (n-4))$ $= 7 - \frac{(n-4)(n-3)}{2}.$ Plug in $n = 12$: $7 - \frac{8 \cdot 9}{2} = -29.$ Thus, A is the correct answer.