T1101 wrote:

What is the remainder when 2^83 is divided by 9?

(A) -4

(B) 0

(C) 4

(D) 5

(E) 8

OA : D

\(2^{83}= 2^2*(2^3)^{27}=2^2*(9-1)^{27}\)

Remainder of \((9-1)^{27}\), when divided by \(9\) would be \(8 \quad (9-1)\).

So \((9-1)^{27}\) can be written as \(9m+8\), where \(m\) is a positive integer.

\(2^2*(9-1)^{27}=2^2*(9m+8)=4*9m+32\).

\(4*9m+32\) or \(32\) divided by \(9\) will leave \(5\) as remainder.

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