Q4

I'm going with C. Could you please confirm what the OA is?

I applied the following values for n=2 & n=10, which gave the remainder of 1 when divided by 3.

Looking at the criteria, options I and II were true using the above mentioned values two values (2 & 10), where as Option III did not work, since it could be a negative or a positive number when you apply the SqRt for 2^n.

I'm wondering if there is a better way to solve this type of problem other than randomly plugging in values for n, like I did.

Please let us know what the OA is. Thank you very much

I will go with C.

When n=0,2,4,6,8,10,... the remainder is 1 if 2^n is divided by 3.
Therefore, II and III are true.

Hmm. OA seems to be E but could be wrong... can anyone confirm?

I think that the answer is D.

Statement 1: N has to be greater than 0. If N=0, we have 1/3 and we cannot have a remainder of 1. In order to have a remainder of 1, 2^N must be greater than 3. - TRUE

Statement 2: It just states that N is even. I couldn't decide if N is always even, I just know that when dividing by 3, we'll have an odd multiple. - NOT SURE

Statement 3: Must be true. We know that N >= 2 in order to have a remainder, in which case any 2^(N/2) will equal an integer. - TRUE

so 1 and 3 are always true....

stmt 1 n= 0 then 2^0 =1 & 1/3 gives remainder as 1

so this is FALSE (n can ne 0)

stmt 2 : 3^n = (-3)^n

now 2^n/3 gives remainder as 1, this is posiible only if n is even coz

2 = 3-1 so 2^n = (3-1)^n here for an even power the 1 will be added to some muliples of 3 giving remainder as 1, whereas for an odd power the 1 will be subracted from some multiples of 3 hence remainder will be 2

so n is even

hence 3^n = (-3)^n TRUE

stmt 3: sqrt(2^n) is an integer TRUE coz n is an even power of 2

Hence Answer should be E

by the way 0 has no sign coz -0=+0=0 so it also nonnegative

That makes sense for statement 1, the remainder for 1/3 is 1 and not non-existent

i agree witin sameer,whats oa ?