Important Points

• As the number is getting divided by 10 the remainder will be the digit in the units place.
• Also the unit place digit repeats as the power increases at fixed intervals

Calculation:

3 ¹ =3                                               3 ⁶ = 72 9

3 ² = 9                                               3 ⁷ = 218 7

3 ³ = 2 7                                             3 ⁸ = 656 1

3 ⁴ = 8 1                                             3 ⁹ = 19683

3 ⁵ = 243

Here if 3 is raised to the powers of 1, 5, 9, 13 .......... the digit at units place = 3

and the powers are following an arithmetic progression with initial term (a ₀) = 1 and common difference (d) = 4.

Therefore, the last term (a _n = a ₀ + (n-1)d) which comes out to be 2017 with n = 505.

So, 3 ²⁰¹⁹  = 3 ²⁰¹⁷\(\times\)  3 ²  = 3 ²⁰¹⁷\(\times\)  9

3 raised to power 4, 8, 12, 16 .... and so on results 1 in the units place So it can be generalized that units place for 3 ⁴ⁿ  = 1

Now, 3 ²⁰¹⁹  = 3 ⁴⁽⁵⁰⁴⁾\(\times\)  3 ³ = 3 ²⁰¹⁶\(\times\)  27

The units place digit in 3 ²⁰¹⁶  = 1 and when multiplied by 7 which is the last digit in 27 we will get 1  \(\times\)  7 = 7.

Hence, option 3 is correct.