Difference between revisions of "2018 AIME I Problems/Problem 9"
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Case 1. | Case 1. | ||
| − | This is probably the simplest: just make a list of possible combinations for <math>{a, b}</math> and <math>{c, d}</math>. We get <math>{1, 15}...{7, 9}</math> for the first and <math>{4, 20}...{11, 13}</math> for the second. That appears to give us <math>7*8=56</math> solutions, right? NO. Because elements can't repeat, take out the supposed sets <math>[a, b, c, d | + | This is probably the simplest: just make a list of possible combinations for <math>{a, b}</math> and <math>{c, d}</math>. We get <math>{1, 15}...{7, 9}</math> for the first and <math>{4, 20}...{11, 13}</math> for the second. That appears to give us <math>7*8=56</math> solutions, right? NO. Because elements can't repeat, take out the supposed sets <math>[a, b, c, d]</math> of <math>{1, 15, 9, 15}, {2, 14, 10, 14}, {3, 13, 11, 13}, {4, 16, 4, 20}, {5, 11, 5, 19}, {5, 11, 11, 13}, {6, 10, 6, 18}, {6, 10, 10, 14}, {7, 9, 9, 15}, {7, 9, 7, 17}</math>. That's ten cases gone. So <math>46</math> for Case 1. |
Case 2. | Case 2. | ||
Revision as of 17:19, 7 March 2018
How many distinct subsets of 
are there such that there are two distinct elements that sum to 
and two distinct elements that sum to 
? Two examples are 
and 
.
Feel free to correct this problem if the wording isn't verbatim (I know it isn't).
Solutions
Solution Exclusion Awareness
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 is absurd.
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 ![$[a, b, c, d]$](http://latex.artofproblemsolving.com/f/b/4/fb4e9295179e0aaee973344e58755a79aa052a8f.png)
of 
. 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 
. Lesson for this problem: Never be scared to attempt an AIME problem. You will oftentimes get it in ~10 minutes.
