

NJJ Inner circle 6439 Posts 
Here is a question for the brainiacs amongst us!
Imagine a 4 x 4 magic square with which you can perform the following standard feat. 1) Circle ANY number and the cross out all the numbers that share the same row or coloumn. (crossing out six numbers) 2) Of the remaining nine numbers, circle one and cross out all the numbers that hare the same row or coloumn (crossing out four numbers) 3) Of the remaining four numbers circle one and cross out all the numbers that share the same row and coloumn. (crossing out two numbers.) 4) Circle the one uncrossed out number. 5) Write down all four digits and add them together. If the square is true, the four digits will ALWAYS add up to the same sum. Here is the question. How many different combinations of four numbers using this system will add up to the predetermined sum? Note: this ONLY includes series of numbers that have had being subjected to the circle/cross system. Rows, coloums, corners, diagonals do not count. 
stanalger Special user St. Louis, MO 996 Posts 
24, the same as the number of hours in most days.
In a 5x5 square, the number of combinations is 120. In an n x n square, the number of combinations is n! 
NJJ Inner circle 6439 Posts 
Wow!
What was your working? Posted: Sep 15, 2005 6:28pm  Wait a minute...there are 16 first selections and for each of those there are nine possible second selections. That is 144 number combinations just for first two! 
stanalger Special user St. Louis, MO 996 Posts 
Here are the 24 possible final results:
OXXX OXXX OXXX OXXX OXXX OXXX XOXX XOXX XXOX XXOX XXXO XXXO XXOX XXXO XOXX XXXO XOXX XXOX XXXO XXOX XXXO XOXX XXOX XOXX XOXX XOXX XOXX XOXX XOXX XOXX OXXX OXXX XXOX XXOX XXXO XXXO XXOX XXXO OXXX XXXO OXXX XXOX XXXO XXOX XXXO OXXX XXOX OXXX XXOX XXOX XXOX XXOX XXOX XXOX OXXX OXXX XOXX XOXX XXXO XXXO XOXX XXXO OXXX XXXO OXXX XOXX XXXO XOXX XXXO OXXX XOXX OXXX XXXO XXXO XXXO XXXO XXXO XXXO OXXX OXXX XOXX XOXX XXOX XXOX XOXX XXOX OXXX XXOX OXXX XOXX XXOX XOXX XXOX OXXX XOXX OXXX 
stanalger Special user St. Louis, MO 996 Posts 
Yes, 16 possibilities for first selection.
9 for second. 4 for third. 1 for fourth. But the order of the selections doesn't matter. So 16*9*4*1 must be divided by 4*3*2*1. 576/24 = 24. Posted: Sep 15, 2005 6:52pm  Easier to think about it this way: The finished result must have four circled numbers. All four of the circled numbers lie in different rows. All four of the circled numbers also lie in different columns. The circled number in the first row can be in any of the FOUR columns. (No other circled numbers can lie in this same column.) Thus the circled number in the second row can be in any of the remaining THREE columns. Two columns have now been eliminated, so the circled number in the third row can be in any of the remaining TWO columns. The circled number in the fourth row can only be in the ONE column not already eliminated. 4*3*2*1=24 
NJJ Inner circle 6439 Posts 
You are so much smarter the me!
Thank you for taking the time to explain it to me. I really appreciate it! Posted: Sep 16, 2005 12:03am  I use Karl Fulves version where you have 4 x rows 4 x coloumns 2 x diagonals 2 x pan diagonals 7 x two by two squares' corners 4 x three by three squares' corners 1 x four by four squares' corners 4 x wrap around squares (any two numbers on and edge + the two numbers on the opposite side of the square) and now 24 x any four numbers that DO NOT share a row or coloumn That gives me a grand total of 52 different ways!!! 
stanalger Special user St. Louis, MO 996 Posts 
Nicholas,
I think you're confusing magic squares with forcing matrices. You can't have a 4 x 4 square that serves as both a magic square AND a forcing matrix. You CAN, however, make a 4 x 4 magic square that totals to the magic constant in 52 different, symmetric ways if the magic constant is an even number. (For odd magic constants, only 36 of the possible 52 patterns will work.) (Also note that both diagonals and both pandiagonals are included in the list of possible end results of the matrix forcing procedure.) 
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