2024 AMC 10A Problem 21

Attempt Problem 21 of the 2024 AMC 10A 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 10A solutions, or check the answer key.

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21.

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)?

[?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: 1990
Small Hint:

A grid whose every row and column is arithmetic has entry f(i,j)=A+Bi+Cj+Dijf(i, j) = A + Bi + Cj + Dij at row i,i, column jj

Big Hint:

Substitute the four known entries into this form and solve for A,B,C,DA, B, C, D

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+22225=29.-10 + 5 + 2 \cdot 22 - 2 \cdot 5 = 29. Thus, C is the correct answer.

Problem 20#20
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