2006 iTest Problems/Problem 6

Problem

What is the remainder when $2^{2006}$ is divided by 7?

$\mathrm{(A)}\,0\quad\mathrm{(B)}\,1\quad\mathrm{(C)}\,2\quad\mathrm{(D)}\,3\quad\mathrm{(E)}\,4\quad\mathrm{(F)}\,5$

Solution

Notice that $2^6 = 64$ , and the remainder when divided by 7 is 1. In other words, $2^6 \equiv 1 \pmod{7}$ . Thus, $2^{6n} \equiv (2^6)^n \equiv 1^n \equiv 1 \pmod{7}$ .

Since $2006 = 6 \cdot 334 + 2$ , we find that $2^{2006} \equiv 2^2 \equiv 4 \pmod{7}$ , so the remainder when $2^{2006}$ gets divided by 7 is $\boxed{\textbf{(E) } 4}$ .

2006 iTest (Problems, Answer Key)
Preceded by: Problem 5	Followed by: Problem 7
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2006 iTest Problems/Problem 6

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