1984 AIME Problems/Problem 4

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

~MRENTHUSIASM

1984 AIME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1984 AIME Problems/Problem 4

Contents

Problem

Solution 1 (Two Variables)

Solution 2 (One Variable)

See also