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Cell Entry Expansion (Posted on 2015-05-30) Difficulty: 3 of 5
Each of the 25 cells in a 5x5 board is zero. Consider any arbitrary cell and increase by 1 the number in this cell and all cells having a common side with it.

Is it possible to obtain the number 2012 in all cells simultaneously?
If so, give an example.
If not, prove it.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution Analytical solution Comment 1 of 1
No, it is not possible for N = 2012.

Mark the cells as follows:

ABFBA
BCDCB
FDEDF
BCDCB
ABFBA

Let's not worry about individual cells.  
Instead, let's try to get a total of N in the E cell, 
   4N in the D cells, 4N in the C cells, 4N in the F cells,
   8N in the B cells and 4N in the A cells.  

We can do this by directly incrementing an A or B or C or D or E cell.
Let a = the total number of times we directly increment an A cell,
    b = the total number of times we directly increment a B cell, etc.
    
From the point of view of an A cell, directly incrementing any one B cell
also indirectly increments an A cell.

So, (1):   a + b = 4N

From the point of view of a B cell, directly incrementing any one A or C or F cell
also indirectly increments 2 B cells.

So, (2):   2a + b + 2c + 2f = 8N

Similarly,
    (3):   b + c + 2d = 4N  (the C cell formula)
    (4):  2c + d + 4e+ f = 4N (the D cell formula)
    (5):   e + d = N  (the E cell formula)
    (6):   b + f + 2d = 4N (the F cell formula)
    
  Subtract (3) from (c), giving c = f
  
  Double (4) and substitute to get
      6c+ 2d + 8e = 8N
   Multiply (5) by 8 to get 8e + 8d = 8N
   Subtract to get d = c = f.
   
 Double (1) to get 2a + 2b = 8N.
 Substitute in (2) to get 2a + b + 4c = 8N
 From these, b = 4c.
 
 Substituting in (3) gives 4N/7 = c = d = f and 16N/7 = b
 
 So, N must be divisible by 7, and N = 2012 fails.
 
 Does any N work?
 
 e = N - d = 3N/7.
 a = 4N - b = 12N/7
 
 The following number of direct increments gives a total of 7 in each cell.  
 
 32123
 21112
 11311
 21112
 32123
 
 So any multiple of 7 works
 
 
   
  
    


  Posted by Steve Herman on 2015-05-30 19:20:15
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