Problem:
The numbers, in order, of each row and the numbers, in order, of each column of a 5×5 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)?
⎣⎢⎢⎢⎢⎢⎡​⋅⋅12⋅⋅​?⋅⋅⋅⋅​⋅⋅⋅16⋅​⋅48⋅⋅⋅​⋅⋅⋅⋅0​⎦⎥⎥⎥⎥⎥⎤​
Answer Choices:
A. 19
B. 24
C. 29
D. 34
E. 39
Solution:
Let aij​ be the integer at row i and column j. It is given that a55​=0,a24​=48, a43​=16, and a31​=12. Suppose a54​=d. Then row 5 is 4d,3d,2d,d,0 because it is an arithmetic progression with common difference −d. The arithmetic progression in column 1 gives
a41​=2a31​+a51​​=212+4d​=6+2d
The arithmetic progression in column 4 gives
a44​=32a54​+a24​​=32d+48​=32​d+16
Row 4 gives
a43​=16=32a44​+a41​​=334​d+32+6+2d​
which implies 48=310​d+38, so d=3. Filling in column 3 with common difference 16−6=10 and column 1 with difference 12−12=0 produces a13​=46 and a11​=12. Finally,
a12​=2a13​+a11​​=246+12​=(C)29​
The full array looks like this:
⎣⎢⎢⎢⎢⎢⎡​1212121212​29​2419149​463626166​634833183​806040200​⎦⎥⎥⎥⎥⎥⎤​
The problems on this page are the property of the MAA's American Mathematics Competitions