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2024 AMC 10A Problem 21

Problem 21 of 25HarderAlgebra

The numbers, in order, of each row and the numbers, in order, of each column of a 5×55 \times 5 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)? [⋅?⋅⋅⋅⋅⋅⋅48⋅12⋅⋅⋅⋅⋅⋅16⋅⋅⋅⋅⋅⋅0]\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

If every row and every column is an arithmetic progression, the entry at row i,i, column jj must take the bilinear form f(i,j)=A+Bi+Cj+Dij.f(i, j) = A + Bi + Cj + Dij. Plug in f(5,5)=0,f(5, 5) = 0, f(2,4)=48,f(2, 4) = 48, f(4,3)=16,f(4, 3) = 16, f(3,1)=12f(3, 1) = 12 and solve: A=−10,A = -10, B=5,B = 5, C=22,C = 22, D=−5.D = -5. So position (1,2)(1, 2) is −10+5+2⋅22−2⋅5=29.-10 + 5 + 2 \cdot 22 - 2 \cdot 5 = 29. Thus, C is the correct answer.
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Tagged: arithmetic sequence · system of equations

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