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

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

@@ Line 1: / Line 1: @@
 == Problem ==
-Let <math>\displaystyle S</math> be a list of [[positive integer]]s - not necessarily [[distinct]] - in which the number <math>\displaystyle 68</math> appears. The [[arithmetic mean]] of the numbers in <math>\displaystyle S</math> is <math>\displaystyle 56</math>. However, if <math>\displaystyle 68</math> is removed, the arithmetic mean of the numbers is <math>\displaystyle 55</math>. What's the largest number that can appear in <math>\displaystyle 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 ==
+== Solution 1 (Two Variables) ==
-Suppose <math>S</math> has <math>n</math> members other than 68, and the sum of these members is <math>s</math>.  Then we're given that <math>\frac{s + 68}{n + 1} = 56</math> and <math>\frac{s}{n} = 55</math>.  Multiplying to clear [[denominator]]s, we have <math>s + 68 = 56n + 56</math> and <math>s = 55n</math> so <math>68 = n + 56</math>, <math>n = 12</math> and <math>s = 12\cdot 55 = 660</math>.  Because the sum and number of the elements of <math>S</math> are fixed, if we want to maximize the largest number in <math>S</math>, we should take all but one member of <math>S</math> to be as small as possible.  Since all members of <math>S</math> are positive integers, the smallest possible value of a member is 1.  Thus the largest possible element is <math>660 - 11 = 649</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
+<cmath>\begin{align*}
+\frac{s+68}{n+1}&=56, \\
+\frac{s}{n}&=55.
+\end{align*}</cmath>
+Clearing denominators, we have
+<cmath>\begin{align*}
+s+68&=56n+56, \\
+s&=55n.
+\end{align*}</cmath>
+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 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>
+~JBL (Solution)
+~MRENTHUSIASM (Reconstruction)
+== Solution 2 (One Variable) ==
+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}
+& & & \\ [-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>
+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
+== Video Solution by OmegaLearn ==
+https://youtu.be/xqo0PgH-h8Y?t=82
+~ pi_is_3.14
 == See also ==
-* [[1984 AIME Problems/Problem 3 | Previous problem]]
+{{AIME box|year=1984|num-b=3|num-a=5}}
-* [[1984 AIME Problems/Problem 5 | Next problem]]
+* [[AIME Problems and Solutions]]
-* [[1984 AIME Problems]]
+* [[American Invitational Mathematics Examination]]
+* [[Mathematics competition resources]]
 [[Category:Intermediate Algebra Problems]]

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