Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 AIME II Problems/Problem 8"
(→‎Solution 4)
 
(13 intermediate revisions by 9 users not shown)
Line 2: Line 2:
 
Find 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>. (The notation <math>\lfloor x\rfloor</math> means the greatest integer less than or equal to <math>x</math>.)
 
Find 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>. (The notation <math>\lfloor x\rfloor</math> means the greatest integer less than or equal to <math>x</math>.)
  
−
== Solution ==
+
== Solutions ==
 
===Solution 1===
 
===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.
  
−
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=44</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:
Line 20: Line 21:
  
 
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>.
 +
 +
====Note====
 +
 +
After getting that <math>\left\lfloor\frac{2002}{45}\right\rfloor=44</math>, for ease of computation above, we can use the fact that <math>(40+k)(49-k)</math> varies solely based on <math>k^2</math> and checking these gives us that the pattern fails at <math>k=0</math> giving us <math>\boxed{049}</math> as the answer.
 +
 +
~Dhillonr25
  
 
===Solution 2===
 
===Solution 2===
Line 28: Line 35:
  
 
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>
 
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>
 +
 +
===Solution 3===
 +
In this solution we use inductive reasoning and a lot of trial and error. Depending on how accurately you can estimate, the solution will come quicker or slower.
 +
 +
Using values of <math>k</math> as <math>1, 2, 3, 4,</math> and <math>5,</math> we can find the corresponding values of <math>n</math> relatively easily. For <math>k = 1</math>, <math>n</math> is in the range <math>[2002-1002]</math>; for <math>k = 2</math>, <math>n</math> is the the range <math>[1001-668]</math>, etc: <math>3, [667,501]; 4, [500-401]; 5, [400-334]</math>. For any positive integer <math>k, n</math> is in a range of <math>\left\lfloor \frac{2002}{k} \right\rfloor -\left\lceil \frac{2002}{k+1} \right\rceil</math>.
 +
 +
Now we try testing <math>k = 1002</math> to get a better understanding of what our solution will look like. Obviously, there will be no solution for <math>n</math>, but we are more interested in how the range will compute to. Using the formula we got above, the range will be <math>1-2</math>. Testing any integer <math>k</math> from <math>1002-2000</math> will result in the same range. Also, notice that each and every one of them have no solution for <math>n</math>. Testing <math>1001</math> gives a range of <math>2-2</math>, and <math>2002</math> gives <math>1-1</math>. They each have a solution for <math>n</math>, and their range is only one value. Therefore, we can assume with relative safety that the integer <math>k</math> we want is the lowest integer that follows this equation
 +
 +
<cmath>\left\lfloor\frac{2002}{k}\right\rfloor + 1 = \left\lceil \frac{2002}{k+1}\right\rceil</cmath>
 +
 +
Now we can easily guess and check starting from <math>k = 1</math>. After a few tests it's not difficult to estimate a few jumps, and it took me only a few minutes to realize the answer was somewhere in the forties (You could also use the fact that <math>45^2=2025</math>). Then it's just a matter of checking them until we get <math>\boxed{049}</math>.
 +
Alternatively, you could use the equation above and proceed with one of the other two solutions listed.
 +
 +
 +
-jackshi2006
 +
 +
Edited and <math>\LaTeX</math>ed by PhunsukhWangdu
 +
 +
===Solution 4===
 +
 +
Here is an intuitive way to approximate the answer is around <math>45</math>: For the function
 +
<math>f(n)=\frac{2002}{n}</math>
 +
its derivative is
 +
<math>-\frac{2002}{n^2}</math>,
 +
which should be close to <math>-1</math> because we need to find the smallest skipped integer. The rest of the steps are the same as Solution 1.
 +
 +
-maxamc
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:22, 14 July 2023

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 $x$.)

Solutions

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 $n=44$. (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 $\boxed{049}$.

Note

After getting that $\left\lfloor\frac{2002}{45}\right\rfloor=44$, for ease of computation above, we can use the fact that $(40+k)(49-k)$ varies solely based on $k^2$ and checking these gives us that the pattern fails at $k=0$ giving us $\boxed{049}$ as the answer.

~Dhillonr25

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 $\boxed{049}.$

Solution 3

In this solution we use inductive reasoning and a lot of trial and error. Depending on how accurately you can estimate, the solution will come quicker or slower.

Using values of $k$ as $1, 2, 3, 4,$ and $5,$ we can find the corresponding values of $n$ relatively easily. For $k = 1$, $n$ is in the range $[2002-1002]$; for $k = 2$, $n$ is the the range $[1001-668]$, etc: $3, [667,501]; 4, [500-401]; 5, [400-334]$. For any positive integer $k, n$ is in a range of $\left\lfloor \frac{2002}{k} \right\rfloor -\left\lceil \frac{2002}{k+1} \right\rceil$.

Now we try testing $k = 1002$ to get a better understanding of what our solution will look like. Obviously, there will be no solution for $n$, but we are more interested in how the range will compute to. Using the formula we got above, the range will be $1-2$. Testing any integer $k$ from $1002-2000$ will result in the same range. Also, notice that each and every one of them have no solution for $n$. Testing $1001$ gives a range of $2-2$, and $2002$ gives $1-1$. They each have a solution for $n$, and their range is only one value. Therefore, we can assume with relative safety that the integer $k$ we want is the lowest integer that follows this equation

\[\left\lfloor\frac{2002}{k}\right\rfloor + 1 = \left\lceil \frac{2002}{k+1}\right\rceil\]

Now we can easily guess and check starting from $k = 1$. After a few tests it's not difficult to estimate a few jumps, and it took me only a few minutes to realize the answer was somewhere in the forties (You could also use the fact that $45^2=2025$). Then it's just a matter of checking them until we get $\boxed{049}$. Alternatively, you could use the equation above and proceed with one of the other two solutions listed.


-jackshi2006

Edited and $\LaTeX$ed by PhunsukhWangdu

Solution 4

Here is an intuitive way to approximate the answer is around $45$: For the function $f(n)=\frac{2002}{n}$ its derivative is $-\frac{2002}{n^2}$, which should be close to $-1$ because we need to find the smallest skipped integer. The rest of the steps are the same as Solution 1.

-maxamc

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_8&oldid=195720"