Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AIME II Problems/Problem 2"
(→‎Solution 1)
Line 19: Line 19:
 
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.
  
−
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>.
+
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

Revision as of 18:43, 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 $5^2+7^2+9^2+9^2 = \boxed{162}$. ~akliu

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AIME_II_Problems/Problem_2&oldid=214947"