Question

An array of numbers is constructed beginning with the numbers $-1, 3, 1$ in the top row. Each adjacent pair of numbers is summed to produce a number in the next row. Each row begins and ends with the numbers $-1$ and $1,$ respectively. The first three rows are shown below.

If the process continues, one of the rows will sum to $12{,}288.$ In that row, what is the third number from the left?

$-29$

$-21$

$-14$

$-8$

$-3$

Solution:

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.

2025 AMC 10A Problem 19

Solution:

Problem 19 in Other Years