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

(Solution)
(Solution)
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== 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  
 
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 + 68 = 56n + 56</math> and  
 +
 
<math>s = 55n</math> so  
 
<math>s = 55n</math> so  
 +
 
<math>68 = n + 56</math>,  
 
<math>68 = n + 56</math>,  
 +
 
<math>n = 12</math> and  
 
<math>n = 12</math> and  
 +
 
<math>s = 12\cdot 55 = 660</math>.   
 
<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>.
 
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>.
  

Revision as of 10:39, 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 (ProblemsAnswer KeyResources)
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
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