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

(Added problem, solution still needed)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
{{solution}}
+
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.
 +
 
 +
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>.)
 +
 
 +
We can now compute:
 +
<cmath>\left\lfloor\frac{2002}{45}\right\rfloor=44 </cmath>
 +
<cmath>\left\lfloor\frac{2002}{44}\right\rfloor=45 </cmath>
 +
<cmath>\left\lfloor\frac{2002}{43}\right\rfloor=46 </cmath>
 +
<cmath>\left\lfloor\frac{2002}{42}\right\rfloor=47 </cmath>
 +
<cmath>\left\lfloor\frac{2002}{41}\right\rfloor=48 </cmath>
 +
<cmath>\left\lfloor\frac{2002}{40}\right\rfloor=50 </cmath>
 +
 
 +
From the observation above (and the fact that <math>\left\lfloor\frac{2002}{2002}\right\rfloor=1</math>) we know that all integers between <math>1</math> and <math>44</math> will be achieved for some values of <math>n</math>. Similarly, for <math>n<40</math> we obviously have <math>\left\lfloor\frac{2002}{n}\right\rfloor > 50</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>.
  
 
== 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 21:01, 28 January 2009

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

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 smallest $n$ 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 $\boxed{049}$.

See also

2002 AIME II (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS