oh btw i managed to answer the Q i had by myself (after lying down on my bed - my fav thinking position )

[spoiler] 2(2N+1) ... 2N always gives an EVEN number (regardless of wat N is)... and then 2N + 1 will always give an ODD number... (E+O=O)... so 2 is not a factor in 2N + 1 (since its odd).. now we take (2N +1) and multiply it with 2.. so obviously the result 2(2N + 1) will be a multiple of 2 but NOT a multiple of 4 (i.e. there is only one "2" in it)... and 3 ^ multiple of 2, will always give us a number with the units digit = 9[/spoiler]

correct me if im wrong

You are right here;

Just identify the pattern here;

3^2 = 9
3^4 = 81
3^6 = 729
3^8 = 6561.

Now every alternate even power ends in 9 and when 1 is added to it would be divisible by 10.

3^(4n + 2) gives us powers of 6,10,14 etc which all end in 9 (irrespective of value of n)

All we need to know now is if m =1?

and statement 2 gives us that.

the same as Shankar suggests - identify the pattern BUT 3^(4n + 2) is actually equal to 9^(2n+1) OR 9*9^(2n). It's clear that for any n you get even number in the powers, hence 9^(2n) will always have the unit's digit of 1 (because of cycles => 9^2=81, 9^4=...1, etc.)

now 9*9^(2n) will return the unit's digit 9. When m=1 we have the new unit's digit=0 and the number divisible by 10.

solved as such and the answer is B.

rohit_gmat wrote:n and m are positive integers, what is remainder when 3^(4n + 2) + m is divided by 10?
a. n = 2
b. m = 1

OA B
[spoiler]Need to know if there is any rule about 2(2N + 1)?[/spoiler]

n and m are positive integers, what is remainder when 3^(4n + 2) + m is divided by 10?

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81

3^5 = 243 and the pattern continues 3^4n always has 1 as the units digit. so, 3^4n+2 ends in 1*9 (3^4n * 3^2), so the question is is 1 + m divisible by 10 ?

a. n = 2

[spoiler]1+m = ? Insufficient[/spoiler]

b. m = 1

[spoiler]1+m = 2 Sufficient[/spoiler]

OptionB

3^(4n+2)+m


last digit of powers of 3 follow the follying pattern:
3^1 => 3
3^2 => 9
3^3 => 7
3^4 => 1
3^5 => 3
3^6 => 9
and so on...

From the above pattern one can conlude that
last digit of
3^4n ==> 1
3^(4n+1) ==> 3
3^(4n+2) ==> 9
3^(4n+3) ==> 7

acc to the ques
3^(4n+2) would have have 9 as ones digit..

Now the remainder of 3^(4n+2)+m when divided by 10 depends only on the value of m

Hense statement b alone is sufficient.