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Difference between revisions of "2019 Mock AMC 10B Problems/Problem 20"
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Case #1: There are <math>2 + 2 + 4 + 4 = 12</math> cases where the <math>6</math> is on the end of a sequence. If so, there is <math>1</math> place where the <math>3</math> cannot go. Since <math>1</math> and <math>5</math> are relatively prime to all available numbers, there is no direct restriction on them. The number of cases <math>= 12( 3 \cdot 2 \cdot 1 - 2) = 48</math>. (<math>-2</math> represents the number of permutations containing <math>3</math> next to <math>6</math>.)
 
Case #1: There are <math>2 + 2 + 4 + 4 = 12</math> cases where the <math>6</math> is on the end of a sequence. If so, there is <math>1</math> place where the <math>3</math> cannot go. Since <math>1</math> and <math>5</math> are relatively prime to all available numbers, there is no direct restriction on them. The number of cases <math>= 12( 3 \cdot 2 \cdot 1 - 2) = 48</math>. (<math>-2</math> represents the number of permutations containing <math>3</math> next to <math>6</math>.)
  
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Case #2: There are <math>2 + 2 + 4 + 4 = 16</math> cases where the <math>6</math> is in the middle of a sequence. If so, there are <math>2</math> places where the <math>3</math> can go. Since <math>1</math> and <math>5</math> are relatively prime to all available numbers, there is no direct restriction on them. The number of cases <math>= 12(3 \cdot 2 \cdot 1 - 4) = 24</math>. (<math>-4</math> represents the number of permutations containing <math>3</math> next to <math>6</math>.)
+
Case #2: There are <math>2 + 2 + 4 + 4 = 12</math> cases where the <math>6</math> is in the middle of a sequence. If so, there are <math>2</math> places where the <math>3</math> can go. Since <math>1</math> and <math>5</math> are relatively prime to all available numbers, there is no direct restriction on them. The number of cases <math>= 12(3 \cdot 2 \cdot 1 - 4) = 24</math>. (<math>-4</math> represents the number of permutations containing <math>3</math> next to <math>6</math>.)
  
 
Therefore, the number of factor-hating permutations <math>= 24 + 48 = \boxed{\text{(E)}72}</math>.
 
Therefore, the number of factor-hating permutations <math>= 24 + 48 = \boxed{\text{(E)}72}</math>.

Latest revision as of 15:21, 8 November 2019

Problem

Define a permutation $a_1a_2a_3a_4a_5a_6$ of the set $1, 2, 3, 4, 5, 6$ to be $\text{ factor-hating }$ if $\text{ gcd }(a_k, a_{k+1}) = 1$ for all $1 \leq k \leq 5$. Find the number of $\text{ factor-hating }$ permutations.

$\textbf{(A) }36 \qquad \textbf{(B) }48 \qquad \textbf{(C) }56 \qquad \textbf{(D) }64 \qquad \textbf{(E) }72 \qquad$

Solution

No even numbers can be neighbors, since they are divisible by $2$. This leaves $4$ possible sequences for the even numbers to occupy: $(a_1, a_3, a_5)$, $(a_2, a_4, a_6)$, $(a_1, a_3, a_6)$, and $(a_1, a_4, a_6)$.

Case #1: There are $2 + 2 + 4 + 4 = 12$ cases where the $6$ is on the end of a sequence. If so, there is $1$ place where the $3$ cannot go. Since $1$ and $5$ are relatively prime to all available numbers, there is no direct restriction on them. The number of cases $= 12( 3 \cdot 2 \cdot 1 - 2) = 48$. ($-2$ represents the number of permutations containing $3$ next to $6$.)

Case #2: There are $2 + 2 + 4 + 4 = 12$ cases where the $6$ is in the middle of a sequence. If so, there are $2$ places where the $3$ can go. Since $1$ and $5$ are relatively prime to all available numbers, there is no direct restriction on them. The number of cases $= 12(3 \cdot 2 \cdot 1 - 4) = 24$. ($-4$ represents the number of permutations containing $3$ next to $6$.)

Therefore, the number of factor-hating permutations $= 24 + 48 = \boxed{\text{(E)}72}$.

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