I used this trick whenever I was teaching addition of positive and negative numbers. After discussing the rules for adding signed numbers, I display a table of 16 numbers (or any square array of numbers). First, I tell the students that I am going to have them pick any four numbers from the following array. I tell my students that my favorite number is 7, and how nice it would be if they would pick four numbers that would add up to my favorite number. The only restriction about picking numbers is that once they have picked a number from a particular row and column, they cannot pick another number from either that row or that column.     2         5         9         4         -5         -2         2         -3         5         8         12         7         -7         -4         0         -5     When a student selects a number from the table, I circle that number and cross out the other numbers in that row and column. Try the trick to make sure that it works. How does it work? The table is really just an addition table. The restriction above forces the student to choose exactly one number from each row and column. So, the sum of the four numbers is really the sum of the 8 numbers of the addition table (the blue letters in the row and column headings below). The sum of a + b + c + d + e + f + g + h = 7.     +         a         b         c         d         e         2         5         9         4         f         -5         -2         2         -3         g         5         8         12         7         h         -7         -4         0         -5     If a = -3, can you determine the values of the other seven numbers in the addition table? Please send in your answer in the following format: b = c = etc.

Solution to the Problem: a = -3 b = 0 c = 4 d = -1 e = 5 f = -2 g = 8 h = -4 Since this is an addition table, and since it is given that a = - 3, a + e = 2     so e = 5 b + e = 5     so b = 0 c + e = 9     so c = 4 d + e = 4     so d = -1 a + f = -5     so f = -2 a + g = 5     so g = 8 a + h = -7     so h = -4     +         -3         0         4         -1         5         2         5         9         4         -2         -5         -2         2         -3         8         5         8         12         7         -4         -7         -4         0         -5

Correctly solved by: