ogmega wrote:
If n is an integer such that n > 9, which of the following could be the remainder when \((2 + 2^2 + 2^3 +2^4 + .... +2^n)\) is divided by 3 ?
I. 0
II. 1
III. 2
(A) Only I
(B) Only II
(C) Only III
(D) Only I and III
(E) I, II and III
Now 3 can leave a remainder as 0, 1 or 2.
But we have to concentrate on power of 2s.\(2^{Odd}.....2^1=2=0+2.......2^3=8=6+2.....2^5=32=30+2\)
So, in every case, the remainder is 2
\(2^{Even}.....2^2=4=3+1.......2^4=16=15+1.....2^6=64=63+1\)
So, in every case, the remainder is 1
Thus, the remainders move in pattern for \(2^1, 2^2, 2^3...\) as \(2,1,2,1,2.....\)
\((2 + 2^2 + 2^3 +2^4 + .... +2^n)\)
So if n is oddsay n=1, then remainder = 2
if n=3, then remainder=2+1+2=5 = 3+2, so 2 again
But if n is evensay n=2, then remainder = 2+1=3, so 0
if n=4, then remainder=2+1+2+1=6, so 0 again
Answer is 0 and 2
D