Difference between revisions of "2018 AIME I Problems/Problem 9"

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==Problem==
 
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.
 
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==
  
==Solution Exclusion Awareness==
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==Solution 1==
 
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>.
 
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>.
  
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.
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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 cannot be true.
  
 
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\}\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
 
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.
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<cmath>\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 12, 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.
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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. Write out all the sets and we get <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.
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Total gives <math>\boxed{210}</math>.
  
 
-expiLnCalc
 
-expiLnCalc
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 +
==Solution 2==
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Let's say our four elements in our subset are <math>a,b,c,d</math>. We have two cases. Note that the order of the elements / the element letters themselves don't matter since they are all on equal grounds at the start.
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<math>\textrm{Case } 1 \textrm{:}</math> <math>a+b = 16</math> and <math>c+d = 24</math>.
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List out possibilities for <math>a+b</math> <math>(\text{i.e. } 1+15, 2+14, 3+13 \text{ etc.})</math> but don't list <math>8+8</math> because those are the same elements and that is restricted.
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Then list out the possibilities for <math>c+d \text{ }(\text{i.e. } 4+20, 5+19, 6+18, \text{ etc.})</math> but don't list <math>12+12</math> because they are the same elements.
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This will give you <math>7 \cdot 8</math> elements, which is <math>56</math>. However, as stated above, we have overlap. Just count starting from <math>a+b \textrm{ - } 15,14,13,4,5,11,6,10,7,9</math> all overlap once, which is <math>10</math>, thus <math>56 - 10 = 46</math> cases in this case. Note that <math>12</math> wasn't included because again, if <math>c+d = 24</math>, <math>c</math> and <math>d</math> cannot be <math>12</math>.
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<math>\textrm{Case } 2 \textrm{:}</math> <math>a+b = 16</math> and <math>b+c = 24</math>.
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Here, <math>b</math> is included in both equations. We can easily see that <math>a, b, c</math> will never equal each other.
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Furthermore, there are 17 choices for <math>d</math> (<math>20 - 3</math> included elements) for each <math>b</math>. Listing out the possible <math>b</math>s, we go from <math>15,14,13,11,10,9,7,6,5,4</math>. Do not include <math>8</math> or <math>12</math> because if they are included, then <math>a/c</math> will be the same as <math>b</math>, which is restricted.
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There are <math>10</math> options there, and thus <math>10 \cdot 17 = 170</math>. But, if <math>a+b = 16</math> and <math>b+c = 24</math>, notice that <math>c-a = 8</math>. That means that if <math>b-d</math> is also <math>8</math>, then we have a double-counted set. Starting with <math>b=15</math>, we have <math>15, 14, 13, 11, 10, 9</math> (where <math>d</math> is <math>7, 6, 5, 3, 2, 1)</math>. That means there are <math>6</math> double-counted cases. Thus <math>170 - 6 = 164</math> cases in this case.
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Adding these up, we get <math>46+164 = \boxed{210}.</math>
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~IronicNinja
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~<math>\LaTeX</math> by AlcBoy1729
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~Formatted by ojaswupadhyay and phoenixfire
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 +
== Solution Python ==
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This code works:
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  int num = 0;
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  for(int i = 1; i <= 20; i++){
 +
    for(int j = i+1; j <= 20; j++){
 +
      for(int k = j+1; k <= 20; k++){
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        for(int m = k+1; m <= 20; m++){
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          if(i+j==16 || i + k == 16 || i + m == 16 || j + k == 16
 +
          || j + m == 16 || k + m == 16){
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            if(i+j==24 || i+k==24 || i+m==24 || j+k==24 || j+m == 24 || k+m==24){
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              num++;
 +
            }
 +
          }
 +
        }
 +
      }
 +
    }
 +
  }
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  cout << num << endl;
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 +
==See Also==
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{{AIME box|year=2018|n=I|num-b=8|num-a=10}}
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{{MAA Notice}}

Revision as of 14:44, 9 November 2020

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.

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 \[\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 12, 4, 20\}, \{5, 11, 5, 19\},\]\[\{5, 11, 11, 13\}, \{6, 10, 6, 18\}, \{6, 10, 10, 14\}, \{7, 9, 9, 15\}, \{7, 9, 7, 17\}\] 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

Let's say our four elements in our subset are $a,b,c,d$. We have two cases. Note that the order of the elements / the element letters themselves don't matter since they are all on equal grounds at the start.


$\textrm{Case } 1 \textrm{:}$ $a+b = 16$ and $c+d = 24$.

List out possibilities for $a+b$ $(\text{i.e. } 1+15, 2+14, 3+13 \text{ etc.})$ but don't list $8+8$ 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 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 $a+b \textrm{ - } 15,14,13,4,5,11,6,10,7,9$ all overlap once, which is $10$, thus $56 - 10 = 46$ cases in this case. Note that $12$ wasn't included because again, if $c+d = 24$, $c$ and $d$ cannot be $12$.


$\textrm{Case } 2 \textrm{:}$ $a+b = 16$ and $b+c = 24$.

Here, $b$ is included 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 thus $10 \cdot 17 = 170$. But, 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. Thus $170 - 6 = 164$ cases in this case.

Adding these up, we get $46+164 = \boxed{210}.$

~IronicNinja ~$\LaTeX$ by AlcBoy1729 ~Formatted by ojaswupadhyay and phoenixfire

Solution Python

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;

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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