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

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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>
 
<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>
  
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==Solution==
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==Solution 1 (Modular Arithmetic)==
  
 
For <math>\textbf{(A)}</math> modulo <math>3,</math>
 
For <math>\textbf{(A)}</math> modulo <math>3,</math>
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\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
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Thus, <math>2^{606} - 1</math> is divisible by <math>3.</math>
 
Thus, <math>2^{606} - 1</math> is divisible by <math>3.</math>
  
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\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
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Thus, <math>2^{606} + 1</math> is divisible by <math>5.</math>
 
Thus, <math>2^{606} + 1</math> is divisible by <math>5.</math>
  
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\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
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Thus, <math>2^{607} + 1</math> is divisible by <math>3.</math>
 
Thus, <math>2^{607} + 1</math> is divisible by <math>3.</math>
  
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\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
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Thus, <math>2^{607} + 3^{607}</math> is divisible by <math>5.</math>
 
Thus, <math>2^{607} + 3^{607}</math> is divisible by <math>5.</math>
  

Revision as of 05:51, 28 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 1 (Modular Arithmetic)

For $\textbf{(A)}$ 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 $\textbf{(B)}$ 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 $\textbf{(D)}$ 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 $\textbf{(E)}$ modulo $5,$ \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 (Problems • Answer Key • Resources)
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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