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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