Any integer that ends with a 6 raised to any power results in an integer that ends with a 6.
So 196213 ends with a 6.
There's that, but the 3 is a little harder.
Notice the last digit of the results of 3 raised to various powers.
When "n" is: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
The last digit of 3n is: 1 3 9 7 1 3 9 7 1 3 9 7 1 3 9 7
That's enough to show that every fourth group keeps repeating 1 3 9 7
So divide 196 by 4 and you get 49 complete groups with no remainder.
That means that 213196 ends with a 7.
6 times 7 is 42 so ............ the last digit of 196213 times 213196 is a 2.
If this is incorrect, then my mistake was starting the table with n = 0.
If the correct way was to start the table with n = 1, then the fourth digit in the repeating group would be 1.
