1987 AIME Problems/Problem 8

Revision as of 15:46, 13 March 2015 by Mathgeek2006 (talk | contribs) (Solution 2)

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 1

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 $n = 112$.

Solution 2

Flip all of the fractions for

\[\begin{array}{ccccc}\frac{15}{8} &>& \frac{k + n}{n} &>& \frac{13}{7}\\  105n &>& 56 (k + n)& >& 104n\\  49n &>& 56k& >& 48n\end{array}\]

Continue as in Solution 1.

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

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