Question

The numbers, in order, of each row and the numbers, in order, of each column of a $5\times5$ array of integers form an arithmetic progression of length $5.$ The numbers in positions $(5,5),\ (2,4),\ (4,3),$ and $(3,1)$ are $0,\ 48,\ 16,$ and $12,$ respectively. What number is in position $(1,2)?$

$\begin{bmatrix} \cdot & ? & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & 48 & \cdot \\ 12 & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & 16 & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & 0 \end{bmatrix}$

$19$

$24$

$29$

$34$

$39$

Solution:

A grid whose rows and columns are all arithmetic has entries of the bilinear form $a(i,j)=\alpha+\beta i+\gamma j+\delta ij.$ The four givens yield $\alpha+5\beta+5\gamma+25\delta=0,\quad \alpha+2\beta+4\gamma+8\delta=48,$ $\alpha+4\beta+3\gamma+12\delta=16,\quad \alpha+3\beta+\gamma+3\delta=12.$

Solving gives $\delta=-5,\ \beta=5,\ \gamma=22,\ \alpha=-10.$ Then $a(1,2)=\alpha+\beta+2\gamma+2\delta=-10+5+44-10=29.$

Thus, the correct answer is C.

2024 AMC 12A Problem 14

Solution:

Problem 14 in Other Years