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Difference between revisions of "2022 AMC 10B Problems/Problem 17"

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Thus, <math>2^{607} + 3^{607}</math> is divisible by 5.
 
Thus, <math>2^{607} + 3^{607}</math> is divisible by 5.
  
−
Therefore, the answer is
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Therefore, the answer is <math>\boxed{\textbf{(C) </math>2^{607} - 1<math>}}</math>.
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\boxed{\textbf{(C) <math>2^{607} - 1</math>}}.
 
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
~MrThinker (LaTeX Error)
  
 
==Video Solution==
 
==Video Solution==

Revision as of 14:32, 17 November 2022

Problem

One of the following numbers is not divisible by any prime number less than 10. Which is it? $\textbf{(A) } 2^{606}-1 \qquad\textbf{(B) } 2^{606}+1 \qquad\textbf{(C) } 2^{607}-1 \qquad\textbf{(D) } 2^{607}+1\qquad\textbf{(E) } 2^{607}+3^{607}$

Solution

For A, modulo 3, \begin{align*} 2^{606} - 1 & \equiv (-1)^{606} - 1 \\ & \equiv 1 - 1 \\ & \equiv 0 . \end{align*}

Thus, $2^{606} - 1$ is divisible by 3.

For B, modulo 5, \begin{align*} 2^{606} + 1 & \equiv 2^{{\rm Rem} ( 606, \phi(5) )} + 1 \\ & \equiv 2^{{\rm Rem} ( 606, 4 )} + 1 \\ & \equiv 2^2 + 1 \\ & \equiv 0 . \end{align*}

Thus, $2^{606} + 1$ is divisible by 5.

For D, modulo 3, \begin{align*} 2^{607} + 1 & \equiv (-1)^{607} + 1 \\ & \equiv - 1 + 1 \\ & \equiv 0 . \end{align*}

Thus, $2^{607} + 1$ is divisible by 3.

For E, module 5, \begin{align*} 2^{607} + 3^{607} & \equiv 2^{607} + (-2)^{607} \\ & \equiv 2^{607} - 2^{607} \\ & \equiv 0 . \end{align*}

Thus, $2^{607} + 3^{607}$ is divisible by 5.

Therefore, the answer is $\boxed{\textbf{(C)$ (Error compiling LaTeX. Unknown error_msg)2^{607} - 1$}}$ (Error compiling LaTeX. Unknown error_msg).

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

~MrThinker (LaTeX Error)

Video Solution

https://youtu.be/YF3HPVcVGZk

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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