Difference between revisions of "2018 AIME I Problems/Problem 9"
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This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set <math>\{a, b, c, d\}</math>. | This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set <math>\{a, b, c, d\}</math>. | ||
− | Note that there are only two cases: 1 where <math>a + b = 16</math> and <math>c + d = 24</math> or 2 where <math>a + b = 16</math> and <math>a + c = 24</math>. Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you <math>a=d</math>, which | + | Note that there are only two cases: 1 where <math>a + b = 16</math> and <math>c + d = 24</math> or 2 where <math>a + b = 16</math> and <math>a + c = 24</math>. Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you <math>a=d</math>, which cannot be true. |
Case 1. | Case 1. |
Revision as of 22:04, 8 March 2018
Find the number of four-element subsets of with the property that two distinct elements of a subset have a sum of , and two distinct elements of a subset have a sum of . For example, and are two such subsets.
Solutions
Solution 1
This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set .
Note that there are only two cases: 1 where and or 2 where and . Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you , which cannot be true.
Case 1. This is probably the simplest: just make a list of possible combinations for and . We get for the first and for the second. That appears to give us solutions, right? NO. Because elements can't repeat, take out the supposed sets That's ten cases gone. So for Case 1.
Case 2. We can look for solutions by listing possible values and filling in the blanks. Start with , as that is the minimum. We find , and likewise up to . But we can't have or because or , respectively! Now, it would seem like there are values for and unique values for each , giving a total of , but that is once again not true because there are some repeated values! We can subtract 1 from all pairs of sets that have two elements in common, because those can give us identical sets. There are 3 pairs about and 3 pairs about , meaning we lose . That's for Case 2.
Total gives .
-expiLnCalc
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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