Difference between revisions of "2006 iTest Problems/Problem 6"

Latest revision as of 23:31, 3 November 2023

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 $2^3 = 8$ , and the remainder when divided by 7 is 1. In other words, $2^3 \equiv 1 \pmod{7}$ . Thus, $2^{3n} \equiv (2^3)^n \equiv 1^n \equiv 1 \pmod{7}$ .

Since $2006 = 3 \cdot 668 + 2$ , 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10

@@ Line 7: / Line 7: @@
 ==Solution==
-Notice that <math>2^6 = 64</math>, and the remainder when divided by 7 is 1.  In other words, <math>2^6 \equiv 1 \pmod{7}</math>.  Thus, <math>2^{6n} \equiv (2^6)^n \equiv 1^n \equiv 1 \pmod{7}</math>.
+Notice that <math>2^3 = 8</math>, and the remainder when divided by 7 is 1.  In other words, <math>2^3 \equiv 1 \pmod{7}</math>.  Thus, <math>2^{3n} \equiv (2^3)^n \equiv 1^n \equiv 1 \pmod{7}</math>.
-Since <math>2006 = 6 \cdot 334 + 2</math>, we find that <math>2^{2006} \equiv 2^2 \equiv 4 \pmod{7}</math>, so the remainder when <math>2^{2006}</math> gets divided by 7 is <math>\boxed{\textbf{(E) } 4}</math>.
+Since <math>2006 = 3 \cdot 668 + 2</math>, we find that <math>2^{2006} \equiv 2^2 \equiv 4 \pmod{7}</math>, so the remainder when <math>2^{2006}</math> gets divided by 7 is <math>\boxed{\textbf{(E) } 4}</math>.
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Difference between revisions of "2006 iTest Problems/Problem 6"

Latest revision as of 23:31, 3 November 2023

Problem

Solution

See Also