Choice E is the answer. You should know the units digits when 3 is raised to a certain power:

Units digit of:
3^1 = 3
3^2 = 9
3^3 = 7
3^4 = 1
.
.
3^5 = 3
3^6 = 9
.
.
. Etc.
So whenever 3 is raised to a certain power, the units digit can only be 3,9,7 or 1, or they repeat after an increase of 4 (i.e - units digit of 3^1 is same as units digit of 3^5 which is 3).

So back to the problem, for any positive integer n, lets say n = 1, then 3 ^ (8 +3) + 2 = 3^11 +2 will have a remainder of 4, since looking back, you can see that 3^11 will have the same units digit as 3^3, which is 7 (remember, 3^11 has same units digit as 3^7 which has same units digit as 3^3).

So 7 + 2 = 9, and when you divide it by 5, you have a remainder of 4. Choice E.

mp2437 wrote:Choice E is the answer. You should know the units digits when 3 is raised to a certain power:

Units digit of:
3^1 = 3
3^2 = 9
3^3 = 7
3^4 = 1
.
.
3^5 = 3
3^6 = 9
.
.
. Etc.
So whenever 3 is raised to a certain power, the units digit can only be 3,9,7 or 1, or they repeat after an increase of 4 (i.e - units digit of 3^1 is same as units digit of 3^5 which is 3).

So back to the problem, for any positive integer n, lets say n = 1, then 3 ^ (8 +3) + 2 = 3^11 +2 will have a remainder of 4, since looking back, you can see that 3^11 will have the same units digit as 3^3, which is 7 (remember, 3^11 has same units digit as 3^7 which has same units digit as 3^3).

So 7 + 2 = 9, and when you divide it by 5, you have a remainder of 4. Choice E.

Hi MP:

I dont understand why u choose an answer based on any number, like n=1. It's not the same for all cases.

You took n=1 as example, and your answer was based on that, so if I take n=2 then 3 ^ (16 +3) + 2 = 3^19 +2 you can see that 3^19 will have 3, so 3+2=5, which makes 3 as units digits, that it, remainder of 3 when divided by 5.
Tha ks
Silvia

I think there is a difference between 3^8n+3 + 2 & 3^(8n+3) + 2.

So what is the question??? If 3^(8n+3) + 2 is the right one then it will be difficult to find out the answer.

Good question Silvia, I saw this post pop up on my radar and forgot that I posted already. Reading the question, I wasn't sure if the parenthesis should be there, and am surprised that I rushed to the solution. In any case, the method I outlined is the correct one.

Now that I look at the problem, it seems the question could be asked 2 ways:

First, what is 3^(8n + 3) + 2 , or second, what is 3^[8(n+3)].

In the first case, it is the same way I mentioned in the previous post. Check your result again for n = 2, you will see that 3^19 is the same as 3^3 which has unit digit of 7, not 3!

However, if the question is like the second form, 3^[8(n+3)] + 2, then you get the following solution:

When you have 3^[8(n+3)], then for any n, you will have a units digit whose value reduces to 1 (say n=1, then 3^(8*4) = 3^32 which = 3^28 = 3^24 = ... = 3^4 = 3^0, which all have units digit of 1).

Then, 1 + 2 = 3, and dividing that by 5 yields a remainder of 3. So it could be choice D.

However, I think the original version means to say the first form, what is remainder when 3^(8n+3) + 2 is divided by 5, in that case it remains choice E.

Make sure to check for algebra! Please let me know if you have any more questions.