Difference between revisions of "1987 AIME Problems/Problem 8"
Mathgeek2006 (talk | contribs) m (→Solution 2) |
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Flip all of the fractions for | Flip all of the fractions for | ||
| − | <cmath>\begin{ | + | <cmath>\begin{array}{ccccc}\frac{15}{8} &>& \frac{k + n}{n} &>& \frac{13}{7}\\ |
| − | 105n > 56 (k + n) > 104n\\ | + | 105n &>& 56 (k + n)& >& 104n\\ |
| − | 49n > 56k > 48n\end{ | + | 49n &>& 56k& >& 48n\end{array}</cmath> |
Continue as in Solution 1. | Continue as in Solution 1. | ||
Revision as of 14:46, 13 March 2015
Contents
Problem
What is the largest positive integer 
for which there is a unique integer 
such that 
?
Solution 1
Multiplying out all of the denominators, we get:
Since 
, 
. Also, 
, so 
. Thus, 
. is unique if it is within a maximum range of 
, so 
.
Solution 2
Flip all of the fractions for
![\[\begin{array}{ccccc}\frac{15}{8} &>& \frac{k + n}{n} &>& \frac{13}{7}\\ 105n &>& 56 (k + n)& >& 104n\\ 49n &>& 56k& >& 48n\end{array}\]](http://latex.artofproblemsolving.com/b/e/f/bef75a0b9482dd0efc4ff0cb17d912353663dc8f.png)
Continue as in Solution 1.
See also
| 1987 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
