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

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~MrThinker (LaTeX Error)
 
~MrThinker (LaTeX Error)
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==Solution 2 (Elimination)==
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Mersenne Primes are primes of the form <math>2^n-1</math>, where <math>n</math> is prime. Using the process of elimination, we can eliminate every option except for <math>\textbf{(A)}</math> and <math>\textbf{(C)}</math>. Clearly, <math>606</math> isn't prime, so the answer must be <math>\boxed{\textbf{(C) }2^{607}-1}</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 16:25, 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)

~MrThinker (LaTeX Error)


Solution 2 (Elimination)

Mersenne Primes are primes of the form $2^n-1$, where $n$ is prime. Using the process of elimination, we can eliminate every option except for $\textbf{(A)}$ and $\textbf{(C)}$. Clearly, $606$ isn't prime, so the answer must be $\boxed{\textbf{(C) }2^{607}-1}$.

Video Solution

https://youtu.be/YF3HPVcVGZk

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

See Also

2022 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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
All AMC 10 Problems and Solutions

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