Solve the Chessboard Average Problem

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In summary, to solve the problem of finding a number that ends in 7, so that when the 7 is moved to the beginning of the number (and the 7 at the end is removed), the resulting number is 5 times larger, we can start by noticing that the last digit of the original number must be 7 and the second to last digit must be 5. Then, by working backwards, we can determine that the number is 142857.
  • #1
jobsism
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Q. Sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the integers on the neighboring squares. (Two squares are neighbours if they share a common edge or vertex. Thus a square can have 8,5 or 3 neighbours depending on its position.) Show that all the sixty four entries are in fact equal.

How does one begin this problem? I initially thought of assuming the contrary(that all the numbers are distinct) and then tried to obtain some ridiculous conclusion. But all I got were equations with innmuerable unknowns! Can anyone please give me a hint on starting the problem? Please DO NOT post the entire solution, as I would like to solve this on my own. Thanks! :D
 
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  • #2
jobsism said:
Q. Sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the integers on the neighboring squares. (Two squares are neighbours if they share a common edge or vertex. Thus a square can have 8,5 or 3 neighbours depending on its position.) Show that all the sixty four entries are in fact equal.

How does one begin this problem? I initially thought of assuming the contrary(that all the numbers are distinct) and then tried to obtain some ridiculous conclusion. But all I got were equations with innmuerable unknowns! Can anyone please give me a hint on starting the problem? Please DO NOT post the entire solution, as I would like to solve this on my own. Thanks! :D
I haven't worked this problem, but I think your plan of assuming the contrary is a good one. Keep in mind however, that the opposite (logical negation) of "all 64 entries are equal" is NOT "all 64 entries are distinct."

The opposite is "not (all 64 entries are distinct)." Another way of saying this is that two or more of the entries are different.
 
  • #3
Pick n to be the minimum of the integers on the square. Now what? This actually seems pretty easy for an olympiad problem.
 
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  • #4
Since this is a finite set of integers, then there must be a greatest integer in the set, call it n. Since n is an average of its surrounding squares, it cannot be strictly greater than every surrounding square, therefore it must be less than or equal to every surrounding square. However since it is the greatest integer in the set, the set must have only 1 integer element, and all squares are equal.
 
  • #5
Ah, so that's how it's done...thanks a lot guys! It seems I'm not Olympiad material after all :D...I was so sure that I could do this on my own, and yet I goofed...
 
  • #6
I've got another one: Find the least number that ends with 7, so that when you put the 7 at the beginning(and removing the 7 at the end), you get a number 5 times larger as that of the original one.
 
  • #7
I'd start by noticing my ten's digit in the original number must be a 0 or a 5 (in order for it to be divisible by 5).

Actually, on second thought, it must be a 5, and can't be a zero.

Also, the leading digit has to be a 1 since if it is a 2 or more, multiplying it by 5 won't give a leading digit of 7.
 
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  • #8
After thinking about it, it was actually easier than I thought. The last digit has to be 7 and the second to last has to be 5 (so the number is at least 57), so 5 x 7 =35 + 50 = 85, so the third to last digit has to be 8, and the number is at least 857. Continuing on in this fashion, then the number is 142857
 

What is the Chessboard Average Problem?

The Chessboard Average Problem is a mathematical puzzle that involves calculating the average of a set of numbers arranged in a specific pattern on a chessboard.

How do you solve the Chessboard Average Problem?

To solve the Chessboard Average Problem, you need to first understand the pattern. Each number on the chessboard is the sum of the two numbers directly above it and to the left. Once you have the pattern, you can use it to calculate the average of the numbers on the chessboard.

Can the Chessboard Average Problem be solved using any size chessboard?

Yes, the Chessboard Average Problem can be solved using any size chessboard. The only requirement is that the numbers on the chessboard follow the given pattern.

Are there any shortcuts or tricks to solving the Chessboard Average Problem?

Yes, there are a few shortcuts that can make solving the Chessboard Average Problem easier. One approach is to use the fact that the average of any set of numbers is equal to the sum of all the numbers divided by the total number of numbers. Another approach is to use the fact that the numbers on the chessboard follow a specific pattern.

What is the significance of the Chessboard Average Problem?

The Chessboard Average Problem is a popular puzzle that helps develop critical thinking and problem-solving skills. It also has applications in various fields such as mathematics, computer science, and economics.

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