Difference between revisions of "2013 AMC 10B Problems/Problem 22"
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Thus, there are <math>16(24) = 384</math> ways for each possible J value. There are <math>3</math> possible J values that still preserve symmetry: <math>384(3) = \boxed{\textbf{(C) }1152}</math> | Thus, there are <math>16(24) = 384</math> ways for each possible J value. There are <math>3</math> possible J values that still preserve symmetry: <math>384(3) = \boxed{\textbf{(C) }1152}</math> | ||
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+ | ==Solution 2== | ||
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+ | As in solution 1, <math>J</math> must be <math>1</math>, <math>5</math>, or <math>9</math> giving us 3 choices. Additionally <math>A+E = B+F = C+G = D+H</math>. This means once we choose <math>J</math> there are <math>8</math> remaining choices. Going clockwise from <math>A</math> we count, <math>8</math> possibilities for <math>A</math>. Choosing <math>A</math> also determines <math>E</math> which leaves <math>6</math> choices for <math>B</math>, once <math>B</math> is chosen it also determines <math>F</math> leaving <math>4</math> choices for <math>C</math>. Once <math>C</math> is chosen it determines <math>G</math> leaving <math>2</math> choices for <math>D</math>. Choosing <math>D</math> determines <math>H</math>, exhausting the numbers. To get the answer we multiply <math>2*4*6*8*3=\boxed{\textbf{(C) }1152}</math>. | ||
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== See also == | == See also == | ||
{{AMC10 box|year=2013|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2013|ab=B|num-b=21|num-a=23}} |
Revision as of 18:48, 3 February 2014
Contents
Problem
The regular octagon has its center at
. Each of the vertices and the center are to be associated with one of the digits
through
, with each digit used once, in such a way that the sums of the numbers on the lines
,
,
, and
are all equal. In how many ways can this be done?
Solution
First of all, note that must be
,
, or
to preserve symmetry. We also notice that
.
WLOG assume that . Thus the pairs of vertices must be
and
,
and
,
and
, and
and
. There are
ways to assign these to the vertices. Furthermore, there are
ways to switch them (i.e. do
instead of
).
Thus, there are ways for each possible J value. There are
possible J values that still preserve symmetry:
Solution 2
As in solution 1, must be
,
, or
giving us 3 choices. Additionally
. This means once we choose
there are
remaining choices. Going clockwise from
we count,
possibilities for
. Choosing
also determines
which leaves
choices for
, once
is chosen it also determines
leaving
choices for
. Once
is chosen it determines
leaving
choices for
. Choosing
determines
, exhausting the numbers. To get the answer we multiply
.
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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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