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Difference between revisions of "2019 Mock AMC 10B Problems/Problem 19"
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<math>3^{1008} + 1 \equiv 3^{504} + 1 \equiv 3^{252} + 1 \equiv 3^{126} + 1 \equiv 3^{63} - 1 \equiv 2</math> <math>\text{mod}</math> <math>4</math>, and <math>3^{63} + 1 \equiv 4</math> <math>\text{mod}</math> <math>8</math>.  
 
<math>3^{1008} + 1 \equiv 3^{504} + 1 \equiv 3^{252} + 1 \equiv 3^{126} + 1 \equiv 3^{63} - 1 \equiv 2</math> <math>\text{mod}</math> <math>4</math>, and <math>3^{63} + 1 \equiv 4</math> <math>\text{mod}</math> <math>8</math>.  
  
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Therefore, the smallest powers of <math>2</math> that divide each of these numbers are <math>2, 2, 2, 2, 2</math>, and <math>4</math>, respectively. The smallest power of <math>2</math> that divides <math>3^{2016} - 1</math> is thus <math>2^5 \cdot 4 = \boxed{\text{(E)} 256}</math>.
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Therefore, the smallest powers of <math>2</math> that divide each of these numbers are <math>2, 2, 2, 2, 2</math>, and <math>4</math>. The smallest power of <math>2</math> that divides <math>3^{2016} - 1</math> is thus <math>2^5 \cdot 4 = \boxed{\text{(E)} 256}</math>.
  
 
<baker77>
 
<baker77>

Revision as of 20:55, 2 November 2019

Problem

What is the largest power of $2$ that divides $3^{2016}-1$?

$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 32 \qquad\textbf{(C)}\ 64 \qquad\textbf{(D)}\ 128 \qquad\textbf{(E)}\ 256$

Solution

$3^{2016} - 1 = (3^{1008} - 1)(3^{1008} + 1) = (3^{504} - 1)(3^{504} + 1)(3^{1008} + 1) = (3^{252} - 1)(3^{252} + 1)(3^{504} + 1)(3^{1008} + 1)$ $= (3^{126} - 1)(3^{126} + 1)(3^{252} + 1)(3^{504} + 1)(3^{1008} + 1) = (3^{63} - 1)(3^{63} + 1)(3^{126} + 1)(3^{252} + 1)(3^{504} + 1)(3^{1008} + 1)$.

By simple mod checking, we find that

$3^{1008} + 1 \equiv 3^{504} + 1 \equiv 3^{252} + 1 \equiv 3^{126} + 1 \equiv 3^{63} - 1 \equiv 2$ $\text{mod}$ $4$, and $3^{63} + 1 \equiv 4$ $\text{mod}$ $8$.

Therefore, the smallest powers of $2$ that divide each of these numbers are $2, 2, 2, 2, 2$, and $4$. The smallest power of $2$ that divides $3^{2016} - 1$ is thus $2^5 \cdot 4 = \boxed{\text{(E)} 256}$.

<baker77>

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