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

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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>?
 
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>?
 
== Solution ==
 
== Solution ==
{{solution}}
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Multiplying out all of the [[denominator]]s, we get:
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:<math>104(n+k) < 195n < 105(n+k)</math>
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:<math>0 < 91n - 104k < n + k</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 112, so <math>n = 112</math>.
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== See also ==
 
== See also ==
* [[1987 AIME Problems]]
 
 
 
{{AIME box|year=1987|num-b=7|num-a=9}}
 
{{AIME box|year=1987|num-b=7|num-a=9}}

Revision as of 14:15, 11 February 2007

Problem

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

Solution

Multiplying out all of the denominators, we get:

$104(n+k) < 195n < 105(n+k)$
$0 < 91n - 104k < n + k$

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