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You just follow the pattern in remainders as increasing powers of 2 are divided by 7.

In this case it's (2,4,1,2,4,1,....)
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The pattern of such questions is rather easy.
There is a whole lot more to these remainder type of questions. Format of such basic questions follow,
are a cyclical pattern.

For example in this question.
2^1 = 2
2^2 = 4
2^3 = 8
2^4 =16
2^5 =32

For such questions,remember one thing,try to get a difference of 1 between numerator and the denominator.

Here the answer is a clear 2.
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Question without an OA:

REM(2^200/7)

By using binomial thrm:

REM ( 2^198 *2^2)/7

REM [ (7+1)^66 * 4] /7

REM[ (1*4)/7]

Hence , 4
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What is the remainder when you divide 2^200 by 7?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

I saw the explanation but I can't how to figure out with this problem in a straightforward manner.....

And in this part I was really confused

The answers follow this pattern:
2 divided by 7 leaves remainder 2................

From Gmatclub test

..............
..............
(2^200)/7
make closest to the denominator. the closest value of 2^something = 8
so, (2^3)66 . 2^2 (bring cube then balancing it, because 200 is not divided by 3)
8^66 . 2^2
= (7+1)^66 . 2^2
just consider the last number of the entire sequence that is (1)^66 . 2^2 = 4 , this is the remainder ,you don't have to divide it by 7. but if you have any negative value then plus that with the divisor.
suppose you got (7-1)^66 . 2^2 = (-1)^66 . 2^2 = -4. Then add it with the divisor you have. -4+7 = 3 is the remainder.
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OPEN DISCUSSION OF THIS QUESTION IS HERE: what-is-the-remainder-when-you-divide-2-200-by-140821.html
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