# 2018 AIME I Problems/Problem 9

## Problem

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.

## 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 $$\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 12, 4, 20\}, \{5, 11, 5, 19\},$$ $$\{5, 11, 11, 13\}, \{6, 10, 6, 18\}, \{6, 10, 10, 14\}, \{7, 9, 9, 15\}, \{7, 9, 7, 17\}$$ 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. Write out all the sets and we get $6$. That's $164$ for Case 2.

Total gives $\boxed{210}$.

-expiLnCalc

## Solution 2

This is gonna be real quick. I considered making this an extension to solution one, but I've decided to make it a separate solution. (Disclaimer: I'm not formatting this properly so if anyone wants to format it correctly using Latex, feel free.)

Let's say our four elements in our subset are $a,b,c,d$. We have two cases.

First, we can have where two distinct elements sum to $16$ and two other distinct elements sum to $24$. Basically, one way of representing this is having $a+b = 16$ and $c+d = 24$. The order of the elements / the element letters themselves don't matter since they are all on equal grounds at the start. List out possibilities for $a+b$ $(\text{i.e. } 1+15, 2+14, 3+13 \text{ etc.})$ but don't include $8+8$ because those are the same elements and that is restricted. Then list out the possibilities for $c+d \text{ }(\text{i.e. } 4+20, 5+19, 6+18, \text{ etc.})$ but again don't list $12+12$ because they are the same elements. This will give you $7 \cdot 8$ elements, which is $56$. However, as stated above, we have overlap. Just count starting from $a+b - 15,14,13,4,5,11,6,10,7,9$ all overlap once, which is $10$, thus $56 - 10 = 46$ cases in this first case. Note that $12$ wasn't included because again, if $c+d = 24$, $c$ and $d$ cannot be $12$.

Our second case is when $a+b = 16$ and $b+c = 24$. Here, one element $(b)$ is included twice in both equations. We can easily see that $a, b, c$ will never equal each other. Furthermore, there are 17 choices for $d (20 - 3$ included elements $)$ for each $b$. Listing out the possible $b$s, we go from $15,14,13,11,10,9,7,6,5,4$. Do not include $8$ or $12$ because if they are included, then $a/c$ will be the same as $b$, which is restricted. There are $10$ options there, and since $10 \cdot 17 = 170$, you would think there are $170$ cases and we are done. That's the tricky part about this problem - not the computation itself, but how easily sillies can be made. This is because, if $a+b = 16$ and $b+c = 24$, notice that $c-a = 8$. That means that if $b-d$ is also $8$, then we have a double-counted set. Starting with $b=15$, we have $15, 14, 13, 11, 10, 9$ (where $d$ is $7, 6, 5, 3, 2, 1)$. That means there are $6$ double-counted cases.

Adding these up, we get $46+170-6 = \boxed{210}.$

~IronicNinja ~ $\LaTeX$ by AlcBoy1729 ~Formatted by ojaswupadhyay $$ Again if anyone wants to format this better feel free, I was just lazy.

## Solution C code

This code works:

 int num = 0;
for(int i = 1; i <= 20; i++){
for(int j = i+1; j <= 20; j++){
for(int k = j+1; k <= 20; k++){
for(int m = k+1; m <= 20; m++){
if(i+j==16 || i + k == 16 || i + m == 16 || j + k == 16
|| j + m == 16 || k + m == 16){
if(i+j==24 || i+k==24 || i+m==24 || j+k==24 || j+m == 24 || k+m==24){
num++;
}
}
}
}
}
}
cout << num << endl;


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