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Difference between revisions of "1984 AIME Problems/Problem 4"
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== Problem ==
 
== Problem ==
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Let <math>S</math> be a list of positive integers - not necessarily distinct - in which the number <math>68</math> appears. The arithmetic mean of the numbers in <math>S</math> is <math>56</math>. However, if <math>68</math> is removed, the arithmetic mean of the numbers is <math>55</math>. What's the largest number that can appear in <math>S</math>?
+
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) ==
 
== Solution 1 (Two Variables) ==
−
Suppose that <math>S</math> has <math>n</math> numbers other than <math>68,</math> and the sum of these numbers is <math>s.</math>
+
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  
 
We are given that  
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s&=55n.
 
s&=55n.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Subtracting the equations, we get <math>68=n+56,</math> from which <math>n=12</math> and <math>s=660.</math>
+
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 is <math>660.</math> To maximize the largest number, we minimize the other eleven numbers:
+
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>
−
 
 
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We can have eleven <math>1</math>s and one <math>660-11\cdot1=\boxed{649}.</math>
 
  
 
~JBL (Solution)
 
~JBL (Solution)
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== Solution 2 (One Variable) ==
 
== Solution 2 (One Variable) ==
−
Suppose that <math>S</math> has <math>n</math> numbers other than <math>68.</math> We have the following table:
+
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}
 
<cmath>\begin{array}{c|c|c|c}
 
& & & \\ [-2.5ex]
 
& & & \\ [-2.5ex]
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\textbf{Final} & n & 55 & 55n
 
\textbf{Final} & n & 55 & 55n
 
\end{array}</cmath>
 
\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
 
~MRENTHUSIASM
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/xqo0PgH-h8Y?t=82
 +
 +
~ pi_is_3.14
  
 
== See also ==
 
== See also ==

Latest revision as of 02:38, 16 January 2023

Problem

Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?

Solution 1 (Two Variables)

Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$

We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$

The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$

~JBL (Solution)

~MRENTHUSIASM (Reconstruction)

Solution 2 (One Variable)

Suppose that $S$ has $n$ numbers other than the $68.$ We have the following table: \[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{Count} & \textbf{Arithmetic Mean} & \textbf{Sum} \\ \hline & & & \\ [-2.5ex] \textbf{Initial} & n+1 & 56 & 56(n+1) \\ \hline & & & \\ [-2.5ex] \textbf{Final} & n & 55 & 55n \end{array}\] We are given that \[56(n+1)-68=55n,\] from which $n=12.$ It follows that the sum of the remaining numbers in $S$ is $55n=660.$ We continue with the last paragraph of Solution 1 to get the answer $\boxed{649}.$

~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
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