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Difference between revisions of "2024 AIME II Problems/Problem 2"

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Say the size is 2. Clearly, this doesn't work as the only list would be <math><cmath>9, 9</cmath></math>, which doesn't satisfy condition 1.
 
Say the size is 2. Clearly, this doesn't work as the only list would be <math><cmath>9, 9</cmath></math>, which doesn't satisfy condition 1.
  
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If the size is 4, then we can have two <math>9</math>s, and a remaining sum of <math>12</math>. Since the other two values in the list must be distinct, and their sum must equal <math>30-18=12</math>, we have that the two numbers are in the form <math>a</math> and <math>12-a</math>. Note that we cannot have both values greater than <math>9</math>, and we cannot have only one value greater than <math>9</math>, because this would make the median <math>9</math>, which violates condition 3. Since the median of the list is a positive integer, this means that the greater of <math>a</math> and <math>12-a</math> must be an odd number. The only valid solution to this is <math>a=5</math>. Thus, our answer is <math>5^2+7^2+9^2+9^2 = \boxed{162}</math>. ~akliu
+
If the size is 4, then we can have two <math>9</math>s, and a remaining sum of <math>12</math>. Since the other two values in the list must be distinct, and their sum must equal <math>30-18=12</math>, we have that the two numbers are in the form <math>a</math> and <math>12-a</math>. Note that we cannot have both values greater than <math>9</math>, and we cannot have only one value greater than <math>9</math>, because this would make the median <math>9</math>, which violates condition 3. Since the median of the list is a positive integer, this means that the greater of <math>a</math> and <math>12-a</math> must be an odd number. The only valid solution to this is <math>a=5</math>. Thus, our answer is <math>5^2+7^2+9^2+9^2 = \boxed{236}</math>. ~akliu
  
 
==Solution 2==
 
==Solution 2==
 
If there were an odd number of elements, the median would be in the set. Thus, we start with 4 elements. For 9 to be the mode, there must be 2 9s. For 9 to not be the median, either both numbers are greater than 9, or both numbers are less than 9. Clearly, both numbers must be less. From here, the numbers are clearly <math>(5,7,9,9)</math>, and we add their squares to get <math>\boxed{236}</math> -westwoodmonster
 
If there were an odd number of elements, the median would be in the set. Thus, we start with 4 elements. For 9 to be the mode, there must be 2 9s. For 9 to not be the median, either both numbers are greater than 9, or both numbers are less than 9. Clearly, both numbers must be less. From here, the numbers are clearly <math>(5,7,9,9)</math>, and we add their squares to get <math>\boxed{236}</math> -westwoodmonster

Revision as of 20:17, 8 February 2024

Problem

A list of positive integers has the following properties:

$\bullet$ The sum of the items in the list is $30$.

$\bullet$ The unique mode of the list is $9$.

$\bullet$ The median of the list is a positive integer that does not appear in the list itself.

Find the sum of the squares of all the items in the list.

Solution 1

The third condition implies that the list's size must be an even number, as if it were an odd number, the median of hte list would surely appear in the list itself.

Therefore, we can casework on what even numbers work.

Say the size is 2. Clearly, this doesn't work as the only list would be $<cmath>9, 9</cmath>$, which doesn't satisfy condition 1.

If the size is 4, then we can have two $9$s, and a remaining sum of $12$. Since the other two values in the list must be distinct, and their sum must equal $30-18=12$, we have that the two numbers are in the form $a$ and $12-a$. Note that we cannot have both values greater than $9$, and we cannot have only one value greater than $9$, because this would make the median $9$, which violates condition 3. Since the median of the list is a positive integer, this means that the greater of $a$ and $12-a$ must be an odd number. The only valid solution to this is $a=5$. Thus, our answer is $5^2+7^2+9^2+9^2 = \boxed{236}$. ~akliu

Solution 2

If there were an odd number of elements, the median would be in the set. Thus, we start with 4 elements. For 9 to be the mode, there must be 2 9s. For 9 to not be the median, either both numbers are greater than 9, or both numbers are less than 9. Clearly, both numbers must be less. From here, the numbers are clearly $(5,7,9,9)$, and we add their squares to get $\boxed{236}$ -westwoodmonster

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