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Difference between revisions of "1984 AIME Problems/Problem 4"

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== Solution 1 (Two Variables) ==
 
== Solution 1 (Two Variables) ==
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Suppose that <math>S</math> has <math>n</math> numbers other than that <math>68,</math> and the sum of these numbers is <math>s.</math>
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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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== Solution 2 (One Variable) ==
 
== Solution 2 (One Variable) ==
−
Suppose that <math>S</math> has <math>n</math> numbers other than that <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]

Revision as of 18:19, 1 July 2021

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

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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