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

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== See also ==
 
== See also ==
 
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{{MAA Notice}}

Revision as of 19:37, 4 July 2013

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 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 $\boxed{049}$.

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 $\boxed{049}.$

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

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