Difference between revisions of "2013 AMC 10B Problems/Problem 22"
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First of all, note that <math>J</math> must be <math>1</math>, <math>5</math>, or <math>9</math> to preserve symmetry. We also notice that <math>A+E = B+F = C+G = D+H</math>. | First of all, note that <math>J</math> must be <math>1</math>, <math>5</math>, or <math>9</math> to preserve symmetry. We also notice that <math>A+E = B+F = C+G = D+H</math>. | ||
− | + | Assume that <math>J = 1</math>. Thus the pairs of vertices must be <math>9</math> and <math>2</math>, <math>8</math> and <math>3</math>, <math>7</math> and <math>4</math>, and <math>6</math> and <math>5</math>. There are <math>4! = 24</math> ways to assign these to the vertices. Furthermore, there are <math>2^{4} = 16</math> ways to switch them (i.e. do <math>2</math> <math>9</math> instead of <math>9</math> <math>2</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> | 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> |
Revision as of 21:08, 6 November 2016
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 1
First of all, note that must be , , or to preserve symmetry. We also notice that .
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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