Difference between revisions of "2002 AIME II Problems/Problem 8"
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== Solution == | == Solution == | ||
+ | ===Solution 1=== | ||
Note that if <math>\frac{2002}n - \frac{2002}{n+1}\leq 1</math>, then either <math>\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor</math>, | Note that if <math>\frac{2002}n - \frac{2002}{n+1}\leq 1</math>, then either <math>\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor</math>, | ||
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. | ||
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Hence the least positive integer <math>k</math> for which the equation <math>\left\lfloor\frac{2002}{n}\right\rfloor=k</math> has no integer solutions for <math>n</math> is <math>\boxed{049}</math>. | Hence the least positive integer <math>k</math> for which the equation <math>\left\lfloor\frac{2002}{n}\right\rfloor=k</math> has no integer solutions for <math>n</math> is <math>\boxed{049}</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | Rewriting the given information and simplifying it a bit, we have | ||
+ | <cmath> \begin{align*} | ||
+ | k \le \frac{2002}{n} < k+1 &\implies \frac{1}{k} \ge \frac{n}{2002} > \frac{1}{k+1}. \\ &\implies \frac{2002}{k} \ge n > \frac{2002}{k+1}. | ||
+ | \end{align*} </cmath> | ||
+ | |||
+ | Now note that in order for there to be no integer solutions to <math>n,</math> we must have <math>\left\lfloor \frac{2002}{k} \right\rfloor = \left\lfloor \frac{2002}{k+1} \right\rfloor.</math> We seek the smallest such <math>k.</math> A bit of experimentation yields that <math>k=49</math> is the smallest solution, as for <math>k=49,</math> it is true that <math>\left\lfloor \frac{2002}{49} \right\rfloor = \left\lfloor \frac{2002}{50} \right\rfloor = 40.</math> Furthermore, <math>k=49</math> is the smallest such case. (If unsure, we could check if the result holds for <math>k=48,</math> and as it turns out, it doesn't.) Therefore, the answer is <math>\boxed{049}.</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=II|num-b=7|num-a=9}} | {{AIME box|year=2002|n=II|num-b=7|num-a=9}} |
Revision as of 17:40, 14 February 2010
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 smallest 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 |