Difference between revisions of "2018 AIME I Problems/Problem 9"
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==Solution Exclusion Awareness== | ==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 <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>. |
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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 is absurd. | 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 is absurd. | ||
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} | + | 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\}\dots\{7, 9\}</math> for the first and <math>\{4, 20\}\dots\{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 |
| + | <cmath>\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 16, 4, 20\}, \{5, 11, 5, 19\},</cmath><cmath>\{5, 11, 11, 13\}, \{6, 10, 6, 18\}, \{6, 10, 10, 14\}, \{7, 9, 9, 15\}, \{7, 9, 7, 17\}</cmath> That's ten cases gone. So <math>46</math> for Case 1. | ||
Case 2. | Case 2. | ||
| − | We can look for solutions by listing possible <math>a</math> values and filling in the blanks. Start with <math>a=4</math>, as that is the minimum. We find <math>{4, 12, 20, ?}</math>, and likewise up to <math>a=15</math>. But we can't have <math>a=8</math> or <math>a=12</math> because <math>a=b</math> or <math>a=c</math>, respectively! Now, it would seem like there are <math>10</math> values for <math>a</math> and <math>17</math> unique values for each <math>?</math>, giving a total of <math>170</math>, 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 <math>a=8</math> and 3 pairs about <math>a=12</math>, meaning we lose <math>6</math>. That's <math>164</math> for Case 2. | + | We can look for solutions by listing possible <math>a</math> values and filling in the blanks. Start with <math>a=4</math>, as that is the minimum. We find <math>\{4, 12, 20, ?\}</math>, and likewise up to <math>a=15</math>. But we can't have <math>a=8</math> or <math>a=12</math> because <math>a=b</math> or <math>a=c</math>, respectively! Now, it would seem like there are <math>10</math> values for <math>a</math> and <math>17</math> unique values for each <math>?</math>, giving a total of <math>170</math>, 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 <math>a=8</math> and 3 pairs about <math>a=12</math>, meaning we lose <math>6</math>. That's <math>164</math> for Case 2. |
Total gives <math>\boxed{210}</math>. Lesson for this problem: Never be scared to attempt an AIME problem. You will oftentimes get it in ~10 minutes. | Total gives <math>\boxed{210}</math>. Lesson for this problem: Never be scared to attempt an AIME problem. You will oftentimes get it in ~10 minutes. | ||
-expiLnCalc | -expiLnCalc | ||
Revision as of 18:56, 7 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 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
![\[\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 16, 4, 20\}, \{5, 11, 5, 19\},\]](http://latex.artofproblemsolving.com/4/a/9/4a935e4e076d965e072e64cc0dac959487372318.png)
![\[\{5, 11, 11, 13\}, \{6, 10, 6, 18\}, \{6, 10, 10, 14\}, \{7, 9, 9, 15\}, \{7, 9, 7, 17\}\]](http://latex.artofproblemsolving.com/6/1/f/61f420cd5557117b1cf09fabdd4b4c4952fb3ab9.png)
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.
-expiLnCalc
