2018 AIME I Problems/Problem 9: Difference between revisions
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Find the number of four-element subsets of <math>\{1,2,3,4,\dots, 20\}</math> with the property that two distinct elements of a subset have a sum of <math>16</math>, and two distinct elements of a subset have a sum of <math>24</math>. For example, <math>\{3,5,13,19\}</math> and <math>\{6,10,20,18\}</math> are two such subsets. | |||
==Solutions== | ==Solutions== | ||
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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 | |||
Revision as of 17:27, 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 ![$[a, b, c, d]$](http://latex.artofproblemsolving.com/f/b/4/fb4e9295179e0aaee973344e58755a79aa052a8f.png)
.
Anyone who sees this page: Please edit my [ to curly braces while I look for LATEX on how to do that. I am a noob at LATEX.
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 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.
-expiLnCalc
