Difference between revisions of "2019 Mock AMC 10B Problems/Problem 20"
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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 numbers in this set, 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 cases with a <math>3</math> next to a <math>6</math>.) | 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 numbers in this set, 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 cases with a <math>3</math> next to a <math>6</math>.) | ||
− | Therefore, the number of factor-hating permutations <math>= 24 + 48 = \boxed{72}</math>. | + | Therefore, the number of factor-hating permutations <math>= 24 + 48 = \boxed{\text{(E)}72}</math>. |
Revision as of 12:05, 4 November 2019
Problem
Define a permutation of the set
to be
if
for all
. Find the number of
permutations.
Solution
No even numbers can be neighbors, since they are divisible by . This leaves
possible sequences for the even numbers to occupy:
,
,
, and
.
Case #1: There are cases where the
is on the end of a sequence. If so, there is
place where the
cannot go. Since
and
are relatively prime to all numbers in this set, there is no direct restriction on them. The number of cases
. (
represents the number of cases with a
next to a
.)
Case #2: There are cases where the
is in the middle of a sequence. If so, there are
places where the
can go. Since
and
are relatively prime to all numbers in this set, there is no direct restriction on them. The number of cases
. (
represents the number of cases with a
next to a
.)
Therefore, the number of factor-hating permutations .