Difference between revisions of "1984 AIME Problems/Problem 4"
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== Problem == | == Problem == | ||
| − | == Solution == | + | Let <math>S</math> be a list of positive integers--not necessarily distinct--in which the number <math>68</math> appears. The average (arithmetic mean) of the numbers in <math>S</math> is <math>56</math>. However, if <math>68</math> is removed, the average of the remaining numbers drops to <math>55</math>. What is the largest number that can appear in <math>S</math>? |
| − | {{ | + | |
| + | == Solution 1 (Two Variables) == | ||
| + | Suppose that <math>S</math> has <math>n</math> numbers other than the <math>68,</math> and the sum of these numbers is <math>s.</math> | ||
| + | |||
| + | We are given that | ||
| + | <cmath>\begin{align*} | ||
| + | \frac{s+68}{n+1}&=56, \\ | ||
| + | \frac{s}{n}&=55. | ||
| + | \end{align*}</cmath> | ||
| + | Clearing denominators, we have | ||
| + | <cmath>\begin{align*} | ||
| + | s+68&=56n+56, \\ | ||
| + | s&=55n. | ||
| + | \end{align*}</cmath> | ||
| + | Subtracting the equations, we get <math>68=n+56,</math> from which <math>n=12.</math> It follows that <math>s=660.</math> | ||
| + | |||
| + | The sum of the twelve remaining numbers in <math>S</math> is <math>660.</math> To maximize the largest number, we minimize the other eleven numbers: We can have eleven <math>1</math>s and one <math>660-11\cdot1=\boxed{649}.</math> | ||
| + | |||
| + | ~JBL (Solution) | ||
| + | |||
| + | ~MRENTHUSIASM (Reconstruction) | ||
| + | |||
| + | == Solution 2 (One Variable) == | ||
| + | Suppose that <math>S</math> has <math>n</math> numbers other than the <math>68.</math> We have the following table: | ||
| + | <cmath>\begin{array}{c|c|c|c} | ||
| + | & & & \\ [-2.5ex] | ||
| + | & \textbf{Count} & \textbf{Arithmetic Mean} & \textbf{Sum} \\ | ||
| + | \hline | ||
| + | & & & \\ [-2.5ex] | ||
| + | \textbf{Initial} & n+1 & 56 & 56(n+1) \\ | ||
| + | \hline | ||
| + | & & & \\ [-2.5ex] | ||
| + | \textbf{Final} & n & 55 & 55n | ||
| + | \end{array}</cmath> | ||
| + | We are given that <cmath>56(n+1)-68=55n,</cmath> from which <math>n=12.</math> It follows that the sum of the remaining numbers in <math>S</math> is <math>55n=660.</math> We continue with the last paragraph of Solution 1 to get the answer <math>\boxed{649}.</math> | ||
| + | |||
| + | ~MRENTHUSIASM | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/xqo0PgH-h8Y?t=82 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
== See also == | == See also == | ||
| − | * [[ | + | {{AIME box|year=1984|num-b=3|num-a=5}} |
| − | * [[ | + | * [[AIME Problems and Solutions]] |
| − | * [[ | + | * [[American Invitational Mathematics Examination]] |
| + | * [[Mathematics competition resources]] | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Latest revision as of 02:38, 16 January 2023
Contents
Problem
Let 
be a list of positive integers--not necessarily distinct--in which the number 
appears. The average (arithmetic mean) of the numbers in is 
. However, if is removed, the average of the remaining numbers drops to 
. What is the largest number that can appear in ?
Solution 1 (Two Variables)
Suppose that has 
numbers other than the 
and the sum of these numbers is 
We are given that
Clearing denominators, we have
Subtracting the equations, we get 
from which 
It follows that 
The sum of the twelve remaining numbers in is 
To maximize the largest number, we minimize the other eleven numbers: We can have eleven 
s and one 
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
Solution 2 (One Variable)
Suppose that has numbers other than the 
We have the following table:
![\[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{Count} & \textbf{Arithmetic Mean} & \textbf{Sum} \\ \hline & & & \\ [-2.5ex] \textbf{Initial} & n+1 & 56 & 56(n+1) \\ \hline & & & \\ [-2.5ex] \textbf{Final} & n & 55 & 55n \end{array}\]](http://latex.artofproblemsolving.com/9/c/e/9cecab6cfff01a5a922301a27ed3ff8700086e9b.png)
We are given that ![\[56(n+1)-68=55n,\]](http://latex.artofproblemsolving.com/3/d/e/3def678b885f8e204c4e17af906c952664814bf1.png)
from which It follows that the sum of the remaining numbers in is 
We continue with the last paragraph of Solution 1 to get the answer 
~MRENTHUSIASM
Video Solution by OmegaLearn
https://youtu.be/xqo0PgH-h8Y?t=82
~ pi_is_3.14
See also
| 1984 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
