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

Revision as of 06:24, 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)

We have $\begin{alignat*}{8} 2^{606}-1 &= 4^{303}-1 &&= (4-1)(4^{302}+4^{301}+4^{300}+\cdots+4^0), \\ 2^{606}+1 &= 4^{303}+1 &&= (4+1)(4^{302}-4^{301}+4^{300}-\cdots+4^0), \\ 2^{607}+1 & &&= (2+1)(2^{606}-2^{605}+2^{604}-\cdots+2^0) \\ 2^{607}+3^{607} & &&= (2+3)[(2^{606})(3^0)-(2^{605})(3^1)+(2^{604})(3^2)-\cdots+(2^0)(3^{606})]. \end{alignat*}$

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

We conclude that $\textbf{(A)}$ is divisible by $3$ , $\textbf{(B)}$ is divisible by $5$ , $\textbf{(D)}$ is divisible by $3$ , and $\textbf{(E)}$ is divisible by $5$ .

~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

@@ Line 65: / Line 65: @@
 Since all of the other choices have been eliminated, we are left with <math>\boxed{\textbf{(C) } 2^{607}-1}</math>.
+We conclude that <math>\textbf{(A)}</math> is divisible by <math>3</math>, <math>\textbf{(B)}</math> is divisible by <math>5</math>,<math>\textbf{(D)}</math> is divisible by <math>3</math>, and <math>\textbf{(E)}</math> is divisible by <math>5</math>.
 ~not_slay
-We conclude that <math>\textbf{(A)}</math> is divisible by <math>3</math>, <math>\textbf{(B)}</math> is divisible by <math>5</math>,<math>\textbf{(D)}</math> is divisible by <math>3</math>, and <math>\textbf{(E)}</math> is divisible by <math>5</math>.
 ==Solution 3 (Elimination)==

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

Page

Toolbox

Search