Hmm. looks like a trick question.
3^(4n+2)/10 will always have a reminder of 9, as long as n is a +ve integer. How?
Rem :3^1/10 = 3
3^2/10 = 9
3^3/10 = 7
3^4/10 = 1
3^5/10 = 3
3^6/10 = 9 ---> 6th
The pattern, 3,9,7,1 repeats for every 4 numbers.
4n+2 will always be = 6, 10, 14 ....
For all of these powers of 3 the reminder when divided by 10 is always be 9.
Hence n is not required!! Hence A is not required => B!!
May be I am smoking something.
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