Arrange nine digits in a 3x3 array such that each different digit that appears appears as many times as its cardinality, so for example, if the digit 3 appears at all, it appears 3 times. Also the eight 3-digit numbers formed from the rows, columns (top to bottom) and two diagonals (also top to bottom) must all be different.
Also, the sum of the nine digits used cannot itself be written using the digits you've included in the array.
What is that sum of these nine digits?
(In reply to
A possible solution by Ady TZIDON)
I wound up getting the same solution and I have done a computer assisted exhaustive search that shows the choice of 4,3,2 is the only possible combination of digits. However I missed the solution initially as I interpreted the final requirement of the total of the digits not being able to be written using the digits used as meaning the total can not use any of the digits, however the only solution found 4 4's 3 3's and 2 2's has a total of 29 which uses a two and thus according to the exact wording of the problem is not a valid solution. I think Charlie meant that the total can not be written using only the digits you've included.
There are only 8 ways to pick digits to fill the grid and they are
9
8,1
7,2
6,3
5,4
6,2,1
5,3,1
4,3,2
now 5,3,1 is eliminated as its total is 35
9, (8,1), (7,2), (6,3), (5,4), and (6,2,1) are all eliminated as there is no way to arrange them in a 3x3 grid such that all 8 3-digit numbers are unique. Thus that only leaves (4,3,2) as Ady pointed out.
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Posted by Daniel
on 2010-05-07 14:05:34 |
re: A possible solution
