4X4 Grid

1     2    3    4
5     6    7    8
9    10  11  12
13  14  15  16

Pick a number in the table. Cross out the numbers in the same row and column as the number you selected. There should be nine numbers left. From those nine numbers, repeat the process three more times so that every number will eventually be crossed out. What is the sum of the four numbers you selected? Will this result always occur? Why or why not?

Check out CAS Musings blog


About benvitalis

math grad - Interest: Number theory
This entry was posted in Math Beauty and tagged . Bookmark the permalink.

3 Responses to 4X4 Grid

  1. Pingback: 4×4 Grid and Extensions | CAS Musings

  2. Husain Alshehhi says:

    The answer is always the same. It is obvious that I can always take one number in each column and one number in each row. Assume that the sum of the numbers is S. Denote f(i) as the number of column of the number chosen in the row i. In your example, if we chose the numbers 4,5,11,14, then f(1) = 4, f(2) = 1, f(3) = 3, f(4) = 2, and A(1,f(1)) + A(2,f(2)) + A(3,f(3)) + A(4,f(4)) = 34.

    Note that A(a,f(a)) + A(c,f(c)) = A(a, f(c)) + A(c, f(a)). The statement means that swapping the column between any two rows does not change the sum, i.e., if I chose 4 in the first row and 5 in the second, I can swap them and chose 1 for the first row and 8 for the second, and the sum does not change.

    Therefore, with enough swapping, S = the number on the diamater: S = 1+6+11+15 = 34.

    The problem can be generalized to nxn matrix.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s