Difference between revisions of "1987 AIME Problems/Problem 8"
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== Solution == | == Solution == | ||
| − | + | == Solution 1== | |
Multiplying out all of the [[denominator]]s, we get: | Multiplying out all of the [[denominator]]s, we get: | ||
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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>. | 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>. | ||
| − | + | == Solution 2== | |
Flip all of the fractions for | Flip all of the fractions for | ||
Revision as of 22:15, 12 March 2012
Problem
What is the largest positive integer 
for which there is a unique integer 
such that 
?
Solution
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{align*}\frac{15}{8} > \frac{k + n}{n} > \frac{13}{7}\\
105n > 56 (k + n) > 104n\\
49n > 56k > 48n\end{align*} (Error compiling LaTeX. Unknown error_msg)
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 | ||
