The last digit of the number 3^{2015} is

- 1
- 3
- 5
- 7

**Solution**

The cyclic frequency of number 3 is 4. Last digit comes in iteration of 4 numbers: 3, 9, 7 and 1.

3^{1} = 3

3^{2} = 9

3^{3} = 27

3^{4} = 81

3^{5} = 243

Cyclic frequency means that the last digit repeats after every fourth power. Last digit of:

3^{4n+1} is 3

3^{4n+2} is 9

3^{4n+3} is 7

3^{4n+4} is 1

Now, you have to find remainder when 2015 is divided by 4. The remainder is 3.

2015 = 4n + 3

Last digit will be 7.

**Correct option is D.**