Difference between revisions of "2002 AIME II Problems/Problem 8"
I like pie (talk | contribs) (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 
for which the equation 
has no integer solutions for 
. (The notation 
means the greatest integer less than or equal to 
.)
Solution
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:
![\[\left\lfloor\frac{2002}{45}\right\rfloor=44\]](http://latex.artofproblemsolving.com/6/3/b/63b7ebc5dc5cf733c56ba067b3a0e471f2930a3a.png)
![\[\left\lfloor\frac{2002}{44}\right\rfloor=45\]](http://latex.artofproblemsolving.com/9/6/4/9645b62aba16644838a82c2b030071bd4b0fe59f.png)
![\[\left\lfloor\frac{2002}{43}\right\rfloor=46\]](http://latex.artofproblemsolving.com/b/2/d/b2df80ec34aa1281c7dacdce0e5d844b05ab5a7e.png)
![\[\left\lfloor\frac{2002}{42}\right\rfloor=47\]](http://latex.artofproblemsolving.com/c/f/e/cfe815b974705fc82acc6f64d75abcf1825885fc.png)
![\[\left\lfloor\frac{2002}{41}\right\rfloor=48\]](http://latex.artofproblemsolving.com/3/b/f/3bff94498722646821072a895ce74e7da4b6f534.png)
![\[\left\lfloor\frac{2002}{40}\right\rfloor=50\]](http://latex.artofproblemsolving.com/6/3/a/63a7f66c9af8f3ac644c1de52da4665777f7b460.png)
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 
.
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 | ||
