This problem has to do wiht the cyclicity of units digit of 3

It cycles as follows : 3 9 7 1 3 9 7 1 etc

Question stem : n and m are positive integers

Stmt I

n=2

3 ^ 4n+2 = 3 ^ 10 which will have units digit of 9

If m is 1 then the remiander when divided by 10 is 0
If m is 2 then the remainader whne divided by 10 is 1

There could be different remainder values based on the different values of m

INSUFF

Stmt II
m=1

WE can take n=1,2,3... so from the cyclicity of units digit of 3 we know the units digit of 3 ^ 4n+2 will always be 9 and wehen added to 1 the resulting number (3^4n+2 + m) willl always have a units digit of 0 therefroe when divided by 10 will leave a remainder of 0

SUFF

Hope I dint miss something here

B)

Yep it is C. Thanks to an earlier post from Cramya, I was a able to solve this. Thank you so much.

In the post Cramya mentioned the following:

32. Quick list of repeatable units digit for 2-12 powers here it is:

Some examples:

2^5 = 32 = Unit is 2.
5^12 = 244140625 =Unit is 5

2 = 2, 4, 8, 6
3 = 3, 9, 7, 1
4 = 4, 6
5 = 5
6 = 6
7 = 7, 9, 3, 1
8 = 8, 4, 2, 6
9 = 9, 1
10 = 0
11 = 1
12 = 2, 4,8, 6


Hope that helps