Difference between revisions of "2013 AMC 10B Problems/Problem 22"
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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. Additionally, there are three possible values for <math>J</math>. To get the answer we multiply <math>2*4*6*8*3=\boxed{\textbf{(C) }1152}</math>. | 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. Additionally, there are three possible values for <math>J</math>. To get the answer we multiply <math>2*4*6*8*3=\boxed{\textbf{(C) }1152}</math>. | ||
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+ | == Note == | ||
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+ | There is no rotational reflectional symmetry in the given figure because the figure is labeled with vertices. | ||
== See also == | == See also == |
Revision as of 15:41, 17 September 2019
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, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have:
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. Additionally, there are three possible values for . To get the answer we multiply .
Note
There is no rotational reflectional symmetry in the given figure because the figure is labeled with vertices.
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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