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)
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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
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
