Someone fills a 6x6 matrix with the numbers from 1 to 36 first across, then down, like:
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36
Now they ask for volunteers to randomly select numbers. The number selected will be circled and the others in the same row and column will be crossed out. Non-crossed out numbers are selected until no more numbers can be chosen. For example, selecting 8 means that 7, 9, 10, 11, 12, 2, 14, 20, 26, and 32, can no longer be chosen.
When all is said and done, the total of the circled numbers is 111.
Can you prove why this is so?
To give credit where due, I first saw this on curiousmath.com
(In reply to
re: that was easy (and thoughts) by Alexis)
Since every end position leaves a single number in each row and column, all the end positions can be found as compositions of row and column swaps from another.
However for rows, a and b, since the affect of a row swap on the total is T'= 6 (a-b) + 6 (b-a) + T = T, the total does not change.
Similarly for columns, c and d, the effect of a column swap on the total is T' = (c - d) + (d-c) +T, the total does not change.
Thus the total does not change between solutions and having found one with T = 101, T=101 for all solutions.
Edited on November 10, 2005, 3:27 am
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Posted by goFish
on 2005-11-10 03:25:46 |
Simpler explanation
