2013 AMC 10B Problems/Problem 22
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 | |
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