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I used a different method than those mentioned above, but came out with the same answer.

In this particular question, there was a pattern to the remainders:
(2^1)/3 = 0 (with a remainder of 2)
(2^2)/3 = 1 (with a remainder of 1)
(2^3)/3 = 2 (with a remainder of 2)
(2^4)/3 = 5 (with a remainder of 1)

...and the pattern continues where the odd powers have remainders = 2 and the even powers have remainders = 1.

Using the above, because 28 is an even power, the remainder will be 1.
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What is the remainder when 2^28 is divided by 3?

A. 1
B. 2
C. 3
D. 4
E. 5

Let’s determine the pattern of remainders when a base of 2 is divided by 3.

2^1 / 3 = 0 remainder 2

2^2 / 3 = 1 remainder 1

2^3 / 3 = 2 remainder 2

2^4 / 3 = 5 remainder 1

We see that when 2 is raised to an even power and divided by 3, the remainder is 1. Thus, the remainder when 2^28 is divided by 3 is 1.

Answer: A
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why is the answer from the usual 2^28 not resulting in the answer.

2^28 = 2^4k = which is 6, so 6/3 remainder is 0

:( what is wrong in the approach ? ( i also understand that 0 isn't in the choices. But I'd like to know why this method isn't right )
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We need to find the remainder when \(2^{28}\) is divided by 3?

To solve this problem we need to find the cycle of remainder of power of 2 when divided by 3

Remainder of \(2^1\) (=2) by 3 = 2
Remainder of \(2^2\) (=4) by 3 = 1
Remainder of \(2^3\) (=8) by 3 = 2
Remainder of \(2^4\) (=16) by 3 = 1


=> Cycle is 2

=> Odd Power remainder is 2
=> Even power remainder is 1
=> Remainder of \(2^{28}\) is divided by 3 = 1


So, Answer will be A
Hope it Helps!

Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem

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