You can determine the number of integers in 3^7146 by doing the following calculation:

3^7146 = 10^y

The value of 'y' rounded up to the next highest integer is the number of integers you are looking for.

3^7147 = [3^(log_3 10)] ^ y = 3^[log_3 10)*y]

So y = 7147/(log_3 10) = 7147/2.059 = 3409.986. Hence 3^7147 has 3410 digits.

As for what the last digit is, if you write out the powers of 3 you will see a pattern where the last digit cycles through 3, 9, 7, 1, ... Since 7147 is divisible by 4 with a remainder of 3, the last digit is 7.