Difference between revisions of "2010 AIME I Problems/Problem 14"

Revision as of 13:18, 17 March 2010

Problem

For each positive integer n, let $f(n) = \sum_{k = 1}^{100} \lfloor log_{10} (kn) \rfloor$ . Find the largest value of n for which $f(n) \le 300$ .

Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ .

Solution

Observe that $f$ is strictly increasing in $n$ . We realize that we need $100$ terms to add up to around $300$ , so we need some sequence of $2$ s, $3$ s, and then $4$ s.

It follows that $n \approx 100$ . Manually checking shows that $f(109) = 300$ and $f(110) > 300$ . Thus, our answer is $\boxed{109}$ .

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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Difference between revisions of "2010 AIME I Problems/Problem 14"

Revision as of 13:18, 17 March 2010

Problem

Solution

See also