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If $2^{2006} - 2006$ divided by 7, the remainder is _____

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answer is A.

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Detailed Solution

$2^{2006} = 2^{2} \cdot 2^{2004} = 4 {[2^{3}]}^{668} = 4 [7 + 1]^{668}$

$= 4 [668_{C_{0}} + 668_{C_{1}} 7^{1} + \dots {.668}_{C_{668}} 7^{668}] - 2006$

$= 4 + 7 [some integer] - 2006$

= $= 7 (I n t e g e r) - 2009 = 7 (i n t e g e r) - 7 (287) = m u l t i p l e o f 7$

$Remainder = 0$

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