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
Hope this helped!
I have a very similar question as a matter of fact.
The only difference is that I am trying to compute the remainder when a_2017 is divided by 45.
I couldn't find where the 44 was implemented in guest's answers, otherwise I would have attempted to solve this problem by plugging in 2017 in to his/her's method.
It is clearly not viable to type all the first 2017 numbers into a calculator :D