Difference between revisions of "2002 AIME II Problems/Problem 8"

Revision as of 22:13, 28 June 2015

Problem

Find the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$ . (The notation $\lfloor x\rfloor$ means the greatest integer less than or equal to $x$ .)

Solution

Solution 1

Note that if $\frac{2002}n - \frac{2002}{n+1}\leq 1$ , then either $\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor$ , or $\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor+1$ . Either way, we won't skip any natural numbers.

The greatest $n$ such that $\frac{2002}n - \frac{2002}{n+1} > 1$ is $n=45$ . (The inequality simplifies to $n(n+1)<2002$ , which is easy to solve by trial, as the solution is obviously $\simeq \sqrt{2002}$ .)

We can now compute: $\left\lfloor\frac{2002}{45}\right\rfloor=44$ $\left\lfloor\frac{2002}{44}\right\rfloor=45$ $\left\lfloor\frac{2002}{43}\right\rfloor=46$ $\left\lfloor\frac{2002}{42}\right\rfloor=47$ $\left\lfloor\frac{2002}{41}\right\rfloor=48$ $\left\lfloor\frac{2002}{40}\right\rfloor=50$

From the observation above (and the fact that $\left\lfloor\frac{2002}{2002}\right\rfloor=1$ ) we know that all integers between $1$ and $44$ will be achieved for some values of $n$ . Similarly, for $n<40$ we obviously have $\left\lfloor\frac{2002}{n}\right\rfloor > 50$ .

Hence the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$ is $\boxed{049}$ .

Solution 2

Rewriting the given information and simplifying it a bit, we have $\begin{align*} k \le \frac{2002}{n} < k+1 &\implies \frac{1}{k} \ge \frac{n}{2002} > \frac{1}{k+1}. \\ &\implies \frac{2002}{k} \ge n > \frac{2002}{k+1}. \end{align*}$

Now note that in order for there to be no integer solutions to $n,$ we must have $\left\lfloor \frac{2002}{k} \right\rfloor = \left\lfloor \frac{2002}{k+1} \right\rfloor.$ We seek the smallest such $k.$ A bit of experimentation yields that $k=49$ is the smallest solution, as for $k=49,$ it is true that $\left\lfloor \frac{2002}{49} \right\rfloor = \left\lfloor \frac{2002}{50} \right\rfloor = 40.$ Furthermore, $k=49$ is the smallest such case. (If unsure, we could check if the result holds for $k=48,$ and as it turns out, it doesn't.) Therefore, the answer is $\boxed{049}.$

@@ Line 7: / Line 7: @@
 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 smallest <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>.)
+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:

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

Page

Toolbox

Search