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

(Created page with "==Problem== One of the following numbers is not divisible by any prime number less than 10. Which is it? ==Solution== For A, modulo 3, <cmath> \begin{align*} 2^{606} - 1 &...")
 
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One of the following numbers is not divisible by any prime number less than 10. Which is it?
 
One of the following numbers is not divisible by any prime number less than 10. Which is it?
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<math>\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}</math>
  
 
==Solution==
 
==Solution==

Revision as of 15:31, 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) $2^{607} - 1$}}.

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

Video Solution

https://youtu.be/YF3HPVcVGZk

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

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