Difference between revisions of "1984 AIME Problems/Problem 4"
MRENTHUSIASM (talk | contribs) m (→Problem) |
MRENTHUSIASM (talk | contribs) m |
||
Line 2: | Line 2: | ||
Let <math>S</math> be a list of positive integers - not necessarily distinct - in which the number <math>68</math> appears. The arithmetic mean of the numbers in <math>S</math> is <math>56</math>. However, if <math>68</math> is removed, the arithmetic mean of the numbers is <math>55</math>. What's the largest number that can appear in <math>S</math>? | Let <math>S</math> be a list of positive integers - not necessarily distinct - in which the number <math>68</math> appears. The arithmetic mean of the numbers in <math>S</math> is <math>56</math>. However, if <math>68</math> is removed, the arithmetic mean of the numbers is <math>55</math>. What's 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 <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> and <math>s=660.</math> | ||
− | <math>s = | + | The sum of the twelve remaining numbers 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) | |
− | |||
− | |||
− | |||
− | |||
== See also == | == See also == |
Revision as of 23:43, 21 June 2021
Problem
Let be a list of positive integers - not necessarily distinct - in which the number appears. The arithmetic mean of the numbers in is . However, if is removed, the arithmetic mean of the numbers is . What's the largest number that can appear in ?
Solution 1 (Two Variables)
Suppose that has numbers other than and the sum of these numbers is
We are given that Clearing denominators, we have Subtracting the equations, we get from which and
The sum of the twelve remaining numbers is To maximize the largest number, we minimize the other eleven numbers: We can have eleven s and one
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
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 |