#2**+2 **

Any integer that ends with a 6 raised to any power results in an integer that ends with a 6.

So 196^{213} 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 3^{n} 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 213^{196} ends with a **7**.

6 times 7 is 42 so ............ the last digit of 196^{213} times 213^{196} 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.

Guest May 20, 2021