2010 AIME I Problems/Problem 14

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 1

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 $\boxed{109}$ .

Solution 2

Because we want the value for which $f(n)=300$ , the average value of the 100 terms of the sequence should be around $3$ . For the value of $\lfloor log_{10} (kn) \rfloor$ to be $3$ , $1000 \le kn < 10000$ . We want kn to be around the middle of that range, and for k to be in the middle of 0 and 100, let $k=50$ , so $50n=\frac{10000+1000}{2}=\frac{11000}{2}=5500$ , and $n = 110$ . $f(110) = 301$ , so we want to lower $n$ . Testing $109$ yields $300$ , so our answer is still $\boxed{109}$ .

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2010 AIME I Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

See also