2025 AMC 10A Problem 19

Problem 19 of 25HarderAlgebra

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 will begin and end 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?

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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.