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=32d+16
Row 4 gives
a43=16=32a44+a41=334d+32+6+2d
which implies 48=310d+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:
⎣⎢⎢⎢⎢⎢⎡1212121212292419149463626166634833183806040200⎦⎥⎥⎥⎥⎥⎤
The problems on this page are the property of the MAA's American Mathematics Competitions