2024 AMC 12A Problem 14

Attempt Problem 14 of the 2024 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12A solutions, or check the answer key.

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14.

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), (2,4), (4,3),(5,5),\ (2,4),\ (4,3), and (3,1)(3,1) are 0, 48, 16,0,\ 48,\ 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}

1919

2424

2929

3434

3939

Answer: C
Concepts:arithmetic sequencesystem of equations
Difficulty rating: 1750
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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