Difference between revisions of "1984 AIME Problems/Problem 4"
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== Problem == | == Problem == | ||
− | {{ | + | 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 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 == | == See also == | ||
− | * [[ | + | {{AIME box|year=1984|num-b=3|num-a=5}} |
− | * [[ | + | * [[AIME Problems and Solutions]] |
− | * [[ | + | * [[American Invitational Mathematics Examination]] |
+ | * [[Mathematics competition resources]] | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 03:38, 16 January 2023
Contents
Problem
Let be a list of positive integers--not necessarily distinct--in which the number
appears. The average (arithmetic mean) of the numbers in
is
. However, if
is removed, the average of the remaining numbers drops to
. What is the largest number that can appear in
?
Solution 1 (Two Variables)
Suppose that has
numbers other than the
and the sum of these numbers is
We are given that
Clearing denominators, we have
Subtracting the equations, we get
from which
It follows that
The sum of the twelve remaining numbers in is
To maximize the largest number, we minimize the other eleven numbers: We can have eleven
s and one
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
Solution 2 (One Variable)
Suppose that has
numbers other than the
We have the following table:
We are given that
from which
It follows that the sum of the remaining numbers in
is
We continue with the last paragraph of Solution 1 to get the answer
~MRENTHUSIASM
Video Solution by OmegaLearn
https://youtu.be/xqo0PgH-h8Y?t=82
~ pi_is_3.14
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 |