1well all the numbers are equal clearly!
On a chessboard, the positive real numbers are entered into the squares in such a way that every number is the arithmetic mean of its neighboring squares . Is this possible? If possible then prove it.
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9 Answers
11Let all the square's value be a. If you take n different no.s from the neighbor then the square's value will be =
n*a/n=a
Since this applies for all the squares ,all the square should have the value a.
21Suppose all the numbers are not equal. Let 'a' be maximum of all the numbers. Let the 'a' be entered in 'n' number of squares.(n<64)
There is at least one square among those 'n' squares which has at least one different number in its neighborhood. The different numbers must be less than 'a'. So ,there average can't be equal to 'a'.
Contradiction!
62The proof uses extremal principle and the answer subhodip has given is perfect...
The same idea, but may be it gives a slightly better visualization of the situation....
We subtract the minimum value from each of the squares.... so there is one (or some) squares which have zero entry.... while the others are all positive....
Now we look at the one square with a zero...
Since the other terms are +ve, the average cant be zero....
So each square (around this square) has to be zero...