2024 AMC 12A Problem 14

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

If every row and every column is arithmetic, the entry has the form

\alpha+\beta i+\gamma j+\delta ij

Substitute the four known entries into that form and solve for

\alpha,\beta,\gamma,\delta

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,

\alpha+2\beta+4\gamma+8\delta=48,

\alpha+4\beta+3\gamma+12\delta=16,

\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.

Problem 14 in other years