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

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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> and <math>s=660.</math>
  
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The sum of the twelve remaining numbers 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>
+
The sum of the twelve remaining numbers 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)
 
~JBL (Solution)
  
 
~MRENTHUSIASM (Reconstruction)
 
~MRENTHUSIASM (Reconstruction)
 +
 +
== Solution 2 (One Variable) ==
 +
Suppose that <math>S</math> has <math>n</math> numbers other than <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>
 +
 +
~MRENTHUSIASM
  
 
== See also ==
 
== See also ==

Revision as of 00:49, 22 June 2021

Problem

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

Solution 1 (Two Variables)

Suppose that $S$ has $n$ numbers other than $68,$ 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 $n=12$ and $s=660.$

The sum of the twelve remaining numbers 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 $68.$ We have the following table: \[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{Count} & \textbf{Arithmetic Mean} & \textbf{Sum} \\ \hline & & & \\ [-2.5ex] \textbf{Initial} & n+1 & 56 & 56(n+1) \\ \hline & & & \\ [-2.5ex] \textbf{Final} & n & 55 & 55n \end{array}\]

~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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