2018 AIME I Problems/Problem 9

Revision as of 22:04, 8 March 2018 by Cocohearts (talk | contribs) (Solution 1)

Find the number of four-element subsets of $\{1,2,3,4,\dots, 20\}$ with the property that two distinct elements of a subset have a sum of $16$, and two distinct elements of a subset have a sum of $24$. For example, $\{3,5,13,19\}$ and $\{6,10,20,18\}$ 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 $\{a, b, c, d\}$.

Note that there are only two cases: 1 where $a + b = 16$ and $c + d = 24$ or 2 where $a + b = 16$ and $a + c = 24$. 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 $a=d$, which cannot be true.

Case 1. This is probably the simplest: just make a list of possible combinations for $\{a, b\}$ and $\{c, d\}$. We get $\{1, 15\}\dots\{7, 9\}$ for the first and $\{4, 20\}\dots\{11, 13\}$ for the second. That appears to give us $7*8=56$ solutions, right? NO. Because elements can't repeat, take out the supposed sets \[\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 16, 4, 20\}, \{5, 11, 5, 19\},\]\[\{5, 11, 11, 13\}, \{6, 10, 6, 18\}, \{6, 10, 10, 14\}, \{7, 9, 9, 15\}, \{7, 9, 7, 17\}\] That's ten cases gone. So $46$ for Case 1.

Case 2. We can look for solutions by listing possible $a$ values and filling in the blanks. Start with $a=4$, as that is the minimum. We find $\{4, 12, 20, ?\}$, and likewise up to $a=15$. But we can't have $a=8$ or $a=12$ because $a=b$ or $a=c$, respectively! Now, it would seem like there are $10$ values for $a$ and $17$ unique values for each $?$, giving a total of $170$, 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 $a=8$ and 3 pairs about $a=12$, meaning we lose $6$. That's $164$ for Case 2.

Total gives $\boxed{210}$.

-expiLnCalc

See Also

2018 AIME I (ProblemsAnswer KeyResources)
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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