## Solution to Puzzle #23: Yet Another Magic Square

….and the winner once again is….Ruchir Godura – He happened to be the only one this time to give a correct answer!

Thanks to everyone who tried the puzzle. Here is a link to the video to explain the answer as explained in the original source (Martin Gardner: Mathematical Puzzles and Diversions. Chapter #2):

http://www.educreations.com/lesson/view/solution-to-puzzle-23/9297945/?s=3Yg01m&ref=link

Also reproducing the answer given by Ruchir in its entirety.

My answer is:

The numbers are arranged such that if you take any 2×2 sub square, the sum of the numbers on the diagonals of the 2×2 sub square are equal. As a direct consequence of this, the sum of the numbers on the  diagonals of any 3×3 or 4×4 or 5×5 sub square are also equal.

When you follow the process of putting coins and eliminating squares in the column and row, you are essentially selecting numbers that can only be on a diagonal relative to each other.

You can reconstruct another such 5×5 square starting with any arbitrary numbers and adding up to any arbitrary number , including 57.

example:

start with a 2×2, sum of diagonals is 5

3 4

1 2

add numbers to make it 3×3, sum of diagonals is 14

3 4 7

1 2 5

5 6 9

add squares to make it 4×4, sum of diagonals is 30

3  4  7 10

1  2  5  8

5  6  9 12

9 10 13 16

add squares to make it 5×5 sum of diagonal is 57

3  4  7 10 14

1  2  5  8 12

5  6  9 12 16

9 10 13 16 20

16 17 20 23 27

You can also go from 4×4 to 5×5 and make the sum of diagonals equal to say, 40

3  4  7 10  6

1  2  5  8  4

5  6  9 12  8

9 10 13 16 12

7  8 11 14 10

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