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

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== See also ==
 
== See also ==
 
{{AIME box|year=1984|num-b=3|num-a=5}}
 
{{AIME box|year=1984|num-b=3|num-a=5}}
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* [[AIME Problems and Solutions]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 13:21, 6 May 2007

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 $660 - 11 = 649$.

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