Difference between revisions of "1984 AIME Problems/Problem 4"
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== Problem == | == Problem == | ||
− | Let <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> | + | 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 | ||
== See also == | == See also == |
Latest revision as of 18:19, 1 July 2021
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
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 |