# 2002 AIME II Problems/Problem 8

## 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 $\boxed{049}$.

### 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

## See also

 2002 AIME II (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 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.

Invalid username
Login to AoPS