Difference between revisions of "2002 AIME II Problems/Problem 8"
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or <math>\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor+1</math>. Either way, we won't skip any natural numbers. | or <math>\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor+1</math>. Either way, we won't skip any natural numbers. | ||
− | The | + | The greatest <math>n</math> such that <math>\frac{2002}n - \frac{2002}{n+1} > 1</math> is <math>n=45</math>. (The inequality simplifies to <math>n(n+1)<2002</math>, which is easy to solve by trial, as the solution is obviously <math>\simeq \sqrt{2002}</math>.) |
We can now compute: | We can now compute: |
Revision as of 23:13, 28 June 2015
Problem
Find the least positive integer for which the equation has no integer solutions for . (The notation means the greatest integer less than or equal to .)
Solution
Solution 1
Note that if , then either , or . Either way, we won't skip any natural numbers.
The greatest such that is . (The inequality simplifies to , which is easy to solve by trial, as the solution is obviously .)
We can now compute:
From the observation above (and the fact that ) we know that all integers between and will be achieved for some values of . Similarly, for we obviously have .
Hence the least positive integer for which the equation has no integer solutions for is .
Solution 2
Rewriting the given information and simplifying it a bit, we have
Now note that in order for there to be no integer solutions to we must have We seek the smallest such A bit of experimentation yields that is the smallest solution, as for it is true that Furthermore, is the smallest such case. (If unsure, we could check if the result holds for and as it turns out, it doesn't.) Therefore, the answer is
See also
2002 AIME II (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.