Master the Magic Square: Tips for Solving the 3x3 Puzzle in Just 4 Hours

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Discussion Overview

The discussion revolves around solving a 3x3 magic square puzzle using the numbers 1 to 9, with the specific condition that the difference between the sums of the top and second rows equals the sum of the third row. Participants share their attempts and strategies for solving the puzzle, including the implications of borrowing numbers from other columns.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in solving the puzzle after several hours, consistently being one number short.
  • Another participant proposes a potential solution of the rows as 846, 327, and 519, noting that their approach involved trial and error and the need for borrowing from the tens column.
  • A different participant challenges the idea of borrowing from another column, suggesting it may not be permissible.
  • One participant argues that if borrowing is not allowed, the problem has no solution, explaining that the structure of the magic square requires certain conditions regarding odd and even numbers that cannot be satisfied without borrowing.
  • Another participant claims that if borrowing is permitted, there are 336 possible solutions to the puzzle.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether borrowing is allowed or necessary for solving the puzzle. There are competing views on the validity of the proposed solutions and the implications of borrowing in the context of the magic square.

Contextual Notes

The discussion highlights limitations related to the assumptions about borrowing and the mathematical properties of odd and even numbers within the constraints of the magic square. The implications of these assumptions remain unresolved.

janvianish
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HI

In a square 3 x 3 using the numbers 1 to 9 once only put the numbers so that:
the numbers on the top row minus the numbers in the 2nd row = the numbers on the 3rd row.

trying this for about 4 hrs and am always 1 number out.
 
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Um.. one possible answer:

846, 327 and 519

It was rather trial and error though, and departing from that I should be getting the first digits to be subtracted requiring borrowing from the tens column.
 
hi thanks for this but i do not think we can borrow from another column
 
Unknown008's neat trick of borrowing from the tens column is the only way to solve this problem. If borrowing is not allowed then the problem has no solution.

In fact, each number in the top row would then have to be the sum of the two numbers below it. But an odd number must be the sum of an odd number and an even number; and an even number is either the sum of two even numbers or the sum of two odd numbers. That means that each column must contain either two odd numbers or no odd numbers. Therefore the total number of odd numbers in the square must be even. But there are five odd numbers in the set 1, ..., 9, and five is not an even number. So there is no possibility to fill the square in the required way (except by using the borrowing trick).
 
If borrowing is allowed, then exhaustive search shows that there are 336 solutions.
 
thanks to all for your help...
 

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