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, it is not possible for N = 2012.
Mark the cells as follows:
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
(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.
So any multiple of 7 works