Difference between revisions of "2024 AIME II Problems/Problem 2"
(→Solution 3) |
(→Solution 3) |
||
| Line 26: | Line 26: | ||
==Solution 3== | ==Solution 3== | ||
| − | We can tell that the amount of integers in the list is an even number, because the median of the list doesn't appear in the list. The mode or the most frequent number in the list is 9, so there is more than one 9. We start with three 9's, because they will be easy to eliminate the cases. The list should look like <cmath>9,9,9,x</cmath> or <cmath>x,9,9,9</cmath> in which both cases are impossible because the median is 9, which shows up in the list. The next case is 2 9's which looks like: <cmath>x,y,9,9</cmath> or <cmath>9,9,x,y</cmath> | + | We can tell that the amount of integers in the list is an even number, because the median of the list doesn't appear in the list. The mode or the most frequent number in the list is 9, so there is more than one 9. We start with three 9's, because they will be easy to eliminate the cases. The list should look like <cmath>9,9,9,x</cmath> or <cmath>x,9,9,9</cmath> in which both cases are impossible because the median is 9, which shows up in the list. The next case is 2 9's which looks like: <cmath>x,y,9,9</cmath> or <cmath>9,9,x,y</cmath> Where <math>x+y=12</math> Because both elements of x and y cannot be greater than 9, the set looks like <cmath>x,y,9,9</cmath> and the only number that satisfy this case is 4,8 and 5,7 not 6 and 6 because 9 is the unique mode. it also states the median is an integer, so the pair is 5 and 7 and the set looks like <cmath>5,7,9,9</cmath> and <cmath>5^2+7^2+9^2+9^2=236</cmath> |
==Video Solution== | ==Video Solution== | ||
Revision as of 10:29, 10 February 2024
Problem
A list of positive integers has the following properties:
The sum of the items in the list is 
.
The unique mode of the list is 
.
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 
, which doesn't satisfy condition 1.
If the size is 4, then we can have two s, and a remaining sum of 
. Since the other two values in the list must be distinct, and their sum must equal 
, we have that the two numbers are in the form 
and 
. Note that we cannot have both values greater than , and we cannot have only one value greater than , because this would make the median , which violates condition 3. Since the median of the list is a positive integer, this means that the greater of and must be an odd number. The only valid solution to this is 
. Thus, our answer is 
. ~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 
, and we add their squares to get 
-westwoodmonster
Solution 3
We can tell that the amount of integers in the list is an even number, because the median of the list doesn't appear in the list. The mode or the most frequent number in the list is 9, so there is more than one 9. We start with three 9's, because they will be easy to eliminate the cases. The list should look like ![\[9,9,9,x\]](http://latex.artofproblemsolving.com/a/0/c/a0c5f53f17499bf3eb6d1e8cd56824f302ad3ad3.png)
or ![\[x,9,9,9\]](http://latex.artofproblemsolving.com/1/d/8/1d834cd45b3d5bb26c224651a1e070d643a7224a.png)
in which both cases are impossible because the median is 9, which shows up in the list. The next case is 2 9's which looks like: ![\[x,y,9,9\]](http://latex.artofproblemsolving.com/e/6/5/e65e21d66391fcda4efff6b82ced7b4a2e6250c1.png)
or ![\[9,9,x,y\]](http://latex.artofproblemsolving.com/4/8/1/4811644b170e05885488f51456f0a8b943e685f6.png)
Where 
Because both elements of x and y cannot be greater than 9, the set looks like and the only number that satisfy this case is 4,8 and 5,7 not 6 and 6 because 9 is the unique mode. it also states the median is an integer, so the pair is 5 and 7 and the set looks like ![\[5,7,9,9\]](http://latex.artofproblemsolving.com/f/7/1/f71cdb0ab7771d95935c9e7777a574c5b4bc8f12.png)
and ![\[5^2+7^2+9^2+9^2=236\]](http://latex.artofproblemsolving.com/c/a/b/cab1093c8de9fcadbbcb45417393f522173cdae8.png)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
| 2024 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
