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The figure below contains seven circles arranged in a hexagonal shape. Each circle is to contain a different integer between 1 and 9, inclusive, such that the sum of the three integers contained within any triangle is divisible by 3. In how many ways can this be done? Two ways that differ only by rotation or reflection are distinguishable.

 

 Jan 5, 2019
 #1
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First let's fill it in with just the numbers 0,1,2, which are the remainders of each number on division by 3. Each triangle must have one of each. Put one of them in the center. You can't use any more of that remainder and have to alternate the other two remainders around the outside hexagon. That gives 3 choices for the center and 2 choices for the first cell, say the rightmost. There are 6 ways to fill the diagram with remainders. Given the remainders, you can choose the center in 3 ways and permute the outside cells in (3!)^2 ways, giving a total of

3⋅2⋅3⋅(3!)^2=648 ways.

 

 

Hope this helped!

 Jan 6, 2019

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