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2002 AIME II Problems/Problem 8

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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 (Problems • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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