The answer is A, or as you put it "1.". When something is divisible by 5, the remainder when divided by five will be 0.

If its not a trick question and you meant to put divided by, not "divisible by". The following is what we can conclude;

[(3^1)+2]/5= 1 + 0/5
[(3^2)+2]/5= 2 + 1/5
[(3^3)+2]/5= 5 + 4/5
[(3^4)+2]/5= 16 + 3/5
[(3^5)+2]/5= 49 + 0/5
[(3^6)+2]/5= 146 + 1/5
[(3^7)+2]/5= 437 + 4/5
[(3^8)+2]/5= 1312 + 3/5
[(3^9)+2]/5= 3937 + 0/5

Obviously from this we can see that there is a pattern of 0, 1, 4, 3. The question tells us that were going to be raising 3 to an odd degree becaues 8n+3 tell us so. N times and even number will make that even. Any even number plus an odd number, 8n+3, will be an odd number. As you can see from the evidence above there are different remainders for different odd values of n. When n=1, remainder is 0. When n=3, remainder equals 4. We can conlude that the, when devided by 5 the number will have a remainder of 0 or 4.

We know that there is a pattern and that the remainder must be 0 or 4. We need to think about what were raising it to and what those number fall on.

When n=1, 8*1+3= 11
n=2, 8*2+3= 19
n=3, 8*3+3= 27

If we devide these numbers by four the remainder will tell us where on the pattern of , 0,1,4,3 we will be at. If the remainder of the power divided by 4 is 1, then the remainder of the number in general will be 0. If the remainder of the power divided by 4 is 2, then the remainder of the number in general will be 1. If the remainder of the power divided by 4 is 3, then the remainder of the number in general will be 4.
So ,

11/4=2+3/4
19/4=4+3/4
27/4=6+3/4

Seeing this we can conclude that the power will always be in the third position in the pattern, thus the remainder of the number in general will always be 4.

pradeepkaushal9518 wrote:if n is a positive integer, what is the remainder when (3^8n+3) + 2 is divisible by 5 ?

1.0
2.1
3.2
4.3
5.4

agree with E-4

3^(8n+3)=3^3*3^8n

3^3=27=25+2 so remainder 2, when divide by 5

if n=1 then
3^8=....1
if n=2 then
3^16=.....1
so the same pattern with n=3,4...
....1=000+1 with remainder 1 when divide by 5

when we multiply remaiders from 3^3 and 3^8n=2*1=2
2=5*0+2 with remaider 2

and sum the remainders left after division
2+2=4
4=5*0+4

Just wanted to put a quicker approach

3^(8n+3)+2=3*9^(4n+1)+2

when we divide 9 by 5 reminder is -1 and -1^(4n+1)=-1 as 4n+1 is odd

so when 3*9^(4n+1)+2 is divided by 5 reminder will be -3+2=-1 or reminder is 5-1=4
Ans option 5