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2024 AMC 12A Problem 14

Problem 14 of 25IntermediateAlgebra

The numbers, in order, of each row and the numbers, in order, of each column of a 5×55\times5 array of integers form an arithmetic progression of length 5.5. The numbers in positions (5,5),(5,5), (2,4),(2,4), (4,3),(4,3), and (3,1)(3,1) are 0,0, 48,48, 16,16, and 12,12, respectively. What number is in position (1,2)?(1,2)? [?4812160] \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}

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Solution

A grid whose rows and columns are all arithmetic has entries of the bilinear form a(i,j)=α+βi+γj+δij.a(i,j)=\alpha+\beta i+\gamma j+\delta ij. The four givens yield α+5β+5γ+25δ=0, \alpha+5\beta+5\gamma+25\delta=0, α+2β+4γ+8δ=48, \alpha+2\beta+4\gamma+8\delta=48, α+4β+3γ+12δ=16, \alpha+4\beta+3\gamma+12\delta=16, α+3β+γ+3δ=12. \alpha+3\beta+\gamma+3\delta=12. Solving gives δ=5, β=5, \delta=-5,\ \beta=5,\ γ=22, α=10.\gamma=22,\ \alpha=-10. Then a(1,2)=α+β+2γ+2δa(1,2)=\alpha+\beta+2\gamma+2\delta =10+5+4410=-10+5+44-10 =29.=29. Thus, the correct answer is C.

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Concepts: arithmetic sequence · system of equations

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.