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# how deal with this problem ???

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how deal with this problem ??? [#permalink]  02 Oct 2011, 11:50
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

2 divided by 7 leaves remainder 2................

From Gmatclub test
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Re: how deal with this problem ??? [#permalink]  02 Oct 2011, 12:11
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Usually all problems which deal with big numbers and their divisibility involve some type of sequences.
See:
2^1=2 The remainder is 2
2^2=4 The remainder is 4
2^3=8 The remainder is 1
2^4=16 The remainder is 2
2^5=32 The remainder is 4
2^6=64 The remainder is 1
2^7=128 The remainder is 2
...
If you could see, divisibility by 7 is connected with divisibility of the power by 3. If the power of 2 has the remainder 2 when it is divided by 3, then the remainder from division 2^x by 7 is 4.

Since 200 has the remainder of 2 after division by 3, the answer is (D)

If you are frustrated about the remainder of the number which is fewer than 7, remember that the remainder is an integer r, where x=7*n+r, and n is also integer.
So, for example 2=0*7+2, so the remainder of 2 when it is divided by 7 is 2.
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Re: how deal with this problem ??? [#permalink]  16 Nov 2011, 15:17
answer is D. this is very similar to the 32^32^32 divided by 7 question.
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Re: how deal with this problem ??? [#permalink]  17 Nov 2011, 07:07
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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Re: how deal with this problem ??? [#permalink]  17 Nov 2011, 09:46
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.
Re: how deal with this problem ???   [#permalink] 17 Nov 2011, 09:46
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