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

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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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== Video Solution by OmegaLearn Using Digit Cycles ==
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https://youtu.be/k4eLhi9wXO8
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~ pi_is_3.14
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== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2022|ab=B|num-b=16|num-a=18}}
 
{{AMC10 box|year=2022|ab=B|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:14, 18 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 (Factoring)

$2^{606}-1=4^{303}-1=(4-1)(4^{302}+4^{301}+4^{300}+\dots+4^0)$. A is divisible by 3.

$2^{606}+1=4^{303}+1=(4+1)(4^{302}-4^{301}+4^{300}-\dots+4^0)$. B is divisible by 5.

$2^{607}+1=(2+1)(2^{606}-2^{605}+2^{604}-\dots+2^0)$. D is divisible by 3.

$2^{607}+3^{607}=(2+3)[(2^{606})(3^0)-(2^{605})(3^1)+(2^{604})(3^2)-\dots+(2^0)(3^{606})]$. E is divisible by 5.

Since all of the other choices have been eliminated, we are left with $\boxed{\textbf{(C)}2^{607}-1}$.

~not_slay

Solution 3 (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)

Video Solution by OmegaLearn Using Digit Cycles

https://youtu.be/k4eLhi9wXO8

~ pi_is_3.14


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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