the subtraction of numbers on chess board is at least 5.

Question

assume we own a regular chess board (8 by 8), now we will randomly write in the slots of the board numbers from 1 to 64 (every number we will write exactly one time), show that the probability that {there are two adjacent squares (which share a rib - not diagonally) that the subtraction between the numbers is those squares is at least 5} = the probability of that equal 1 (or 100%).

means that need to show that there are at least one adjacent pair (row or column , not diagonally), so the subtraction of the numbers equal at least 5.

Answer 1

I can do even better: There must be at least one place where the difference is at least $6$.

For contradiction, let's say we have distributed the numbers so that adjacent squares have a difference of at most $5$, and to help in visualising, take a piece that can move like a king, but not diagonally (a so-called wazir). Then the rules can be restated as "There is no place on the board where the wazir can move between two squares whose numbers differ by more than $5$."

That means that numbers that numbers with a big difference must be placed on squares that are many wazir moves apart. The biggest limitation on any pair of numbers is $13$ moves, which corresponds to starting in one corner and ending up one move away from the opposite corner (the biggest separation you can get between two squares is $14$ moves).

Look at all pairs of numbers that must be separated by at least $13$ wazir moves. Those are $$ 1-64\\ 2-64\\ 3-64\\ 1-63\\ 2-63\\ 1-62 $$ That means that on the three squares in one corner we must have $1, 2, 3$, and on the three squares in the opposite corner we must have $62, 63, 64$. Clearly, $1$ and $64$ must go in the actual corner squares, since each of them must be separated by $13$ moves to all the three numbers on the opposite side. However, we cannot place the rest of them in any satisfactory manner (no matter what, $2$ and $63$ will be separated by $12$ moves, for instance).

Thus we see that it is impossible to distribute the numbers in a satisfactory manner deliberately, which means that if you distribute them randomly, you can't possibly manage to do it either, so the probability of it happening is $0$.

Answer 2

First lets look on the place of the digit 1 and the place of the digit 64 on the board. we shell pick a monotonic walk (such a walk that if i went down once i cant go up anymore and otherwise, also if i went left once i cant go right anymore and conversed). now we will use the pigeonhole principle (for the general case),

Every edge will be a "hole" and every subtracting point (by absolute value) between two following squares will be a "pigeon". there for in the walk there is at least 64 - 1 = 63 pigeons and 7+7=14 holes for most (maximum length of a monotonic walk) and from the pigeonhole principle there i a hole with at least $\left \lceil \frac{63}{14} \right \rceil = 5 $ pigeons , means that there is two adjust squares that the subtraction between them is at least 5 !

Answer 3

I am going to prove the stronger:

Claim:

When you put different whole numbers on a $7 \times 7$ board there has to be a pair of adjacent squares with a difference of at least $5$.

Proof:

Proof by contradiction: Suppose there is no such adjacent pair of squares, i.e all adjacent squares have a difference of at most $4$.

Let $n$ be the number in the center square.

Given that adjacent squares have a difference of at most $4$, that means that the numbers in the squares of the sub-board can at most differ by $24$ from $n$, as all squares are within $6$ 'steps' from the center square. And since there are $49$ squares on the board, that means that the numbers on the board have to be $n-24$ through $n+24$.

But while the $4$ squares at the corners of the sub-board can differ by up to $24$ from $n$, the other squares can differ by at most $20$, and hence there is no way to place the $8$ numbers $n-24$, $n-23$, $n-22$, $n-21$, $n+21$, $n+22$, $n+23$, and $n+24$ on the sub-board. Contradiction!

Obviously, if we can't place different whole numbers on a $7 \times 7$ board with adjacent squares having a difference of at most $4$, then we can;t do it for a $8 \times 8$ board either, since if we could do it for an $8 \times 8$ board, then it would be done for any of the $7 \times 7$ 'sub-boards' as well, which we proved is impossible.