Difference between revisions of "1987 AIME Problems/Problem 8"

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== Problem ==
 
== Problem ==
What is the largest positive integer <math>\displaystyle n</math> for which there is a unique integer <math>\displaystyle k</math> such that <math>\displaystyle \frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>?
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What is the largest positive integer <math>n</math> for which there is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>?
 
== Solution ==
 
== Solution ==
 
Multiplying out all of the [[denominator]]s, we get:
 
Multiplying out all of the [[denominator]]s, we get:
  
:<math>104(n+k) < 195n < 105(n+k)</math>
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<cmath>\begin{align*}104(n+k) &< 195n< 105(n+k)\\
:<math>0 < 91n - 104k < n + k</math>
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0 &< 91n - 104k < n + k\end{align*}</cmath>
  
Since <math>91n - 104k < n + k</math>, <math>k > \frac{6}{7}n</math>. Also, <math>0 < 91n - 104k</math>, so <math>k < \frac{7n}{8}</math>. Thus, <math>48n < 56k < 49n</math>. <math>k</math> is unique if it is within a maximum [[range]] of 112, so <math>n = 112</math>.
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Since <math>91n - 104k < n + k</math>, <math>k > \frac{6}{7}n</math>. Also, <math>0 < 91n - 104k</math>, so <math>k < \frac{7n}{8}</math>. Thus, <math>48n < 56k < 49n</math>. <math>k</math> is unique if it is within a maximum [[range]] of <math>112</math>, so <math>n = 112</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1987|num-b=7|num-a=9}}
 
{{AIME box|year=1987|num-b=7|num-a=9}}
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[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Number Theory Problems]]

Revision as of 21:38, 9 April 2008

Problem

What is the largest positive integer $n$ for which there is a unique integer $k$ such that $\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}$?

Solution

Multiplying out all of the denominators, we get:

\begin{align*}104(n+k) &< 195n< 105(n+k)\\ 0 &< 91n - 104k < n + k\end{align*}

Since $91n - 104k < n + k$, $k > \frac{6}{7}n$. Also, $0 < 91n - 104k$, so $k < \frac{7n}{8}$. Thus, $48n < 56k < 49n$. $k$ is unique if it is within a maximum range of $112$, so $n = 112$.

See also

1987 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions