PS

When dividing a number by 5 you are actually looking for its units digit: numbers that end in 0 and 5 are divisible by 5, while the remainder of other numbers can be determined. Say your number ends in 3: this means that the remainder will be 3. But if the number ends in 9, the remainder will be 9 - 5 = 4.

Now, there is a "standard" when dealing with the powers of 3: notice that
3^1 = 3 - units digit 3
3^2 = 9 - units digit 9
3^3 = 27 - units digit 7
3^4 = 81 - units digit 1
3^5 = 243 - units digit 3 and the pattern is repeated every four powers. This means that the units digit of 3^n depends on the nature of n:
n = 4k + 1 - units digit 3
n = 4k + 2 - units digit 9
n = 4k + 3 - units digit 7
n = 4k - units digit 1.

Once you understand this, the problem is a breeze: 8n + 3 = 2*4n + 3, meaning that the units digit of 3^(8n + 3) = 7. Add 2 to that and you get that the units digit of 3^(8n + 3) + 2 = 9. Now, I;m getting remainder 4. Can you please check the way your wrote the problem? I;m pretty sure this is how it;s solved.

then i doubt the OA coz i also got the same answer by the same method.....
Anyways thanks for your reply.

Dana explained well, and that is the way to solve this.
The remainder is 4 for the above problem unless there is a typo in the problem itself.