AIME 1984 · 第 4 题

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

$$\begin{aligned} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{aligned}$$ Clearing denominators, we have

$$\begin{aligned} s+68&=56n+56, \\ s&=55n. \end{aligned}$$ 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