Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution: answer was incorrect)
(→‎Solution)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
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 = \boxed{649}</math>.
+
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 = \boxed{649}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 09:39, 4 August 2017

Problem

Let $\displaystyle S$ be a list of positive integers - not necessarily distinct - in which the number $\displaystyle 68$ appears. The arithmetic mean of the numbers in $\displaystyle S$ is $\displaystyle 56$. However, if $\displaystyle 68$ is removed, the arithmetic mean of the numbers is $\displaystyle 55$. What's the largest number that can appear in $\displaystyle S$?

Solution

Suppose $S$ has $n$ members other than 68, and the sum of these members is $s$. Then we're given that $\frac{s + 68}{n + 1} = 56$ and $\frac{s}{n} = 55$. Multiplying to clear denominators, we have $s + 68 = 56n + 56$ and $s = 55n$ so $68 = n + 56$, $n = 12$ and $s = 12\cdot 55 = 660$. Because the sum and number of the elements of $S$ are fixed, if we want to maximize the largest number in $S$, we should take all but one member of $S$ to be as small as possible. Since all members of $S$ are positive integers, the smallest possible value of a member is 1. Thus the largest possible element is $660 - 11 = \boxed{649}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_4&oldid=86809"