# Difference between revisions of "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

### 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 $\boxed{109}$.

### Solution 3

For any $n$ where the sum is close to $300$, all the terms in the sum must be equal to $2$, $3$ or $4$. Let $M$ be the number of terms less than or equal to $3$ and $N$ be the number of terms equal to $2$ (also counted in $M$). With this definition of $M$ and $N$ the total will be $400 - M - N \le 300$, from which $M + N \ge 100$. Now $M+1$ is the smallest integer $k$ for which $\log_{10}(kn) \ge 4$ or $kn \ge 10000$, thus $$M = \left\lfloor\frac{9999}{n}\right\rfloor.$$ Similarly, $$N = \left\lfloor\frac{999}{n}\right\rfloor = \left\lfloor\frac{M}{10}\right\rfloor.$$

Therefore, $$M + \left\lfloor \frac{M}{10} \right\rfloor \ge 100 \implies M \ge \left\lceil\frac{1000}{11}\right\rceil = 91 \implies n \le \left\lfloor\frac{9999}{91}\right\rfloor = 109.$$ Since we want the largest possible $n$, the answer is $\boxed{109}$.

### Solution 4 (similar to Solution 1, but with more detail)

Since we're working with base-$10$ logarithms, we can start by testing out $n$'s that are powers of $10$. For $n = 1$, the terms in the sum are $\lfloor \log_{10} (1)\rfloor, \lfloor \log_{10} (2)\rfloor, \lfloor \log_{10} (3) \rfloor , . . . , \lfloor \log_{10} (100) \rfloor$. For numbers $1$-$9$, $\lfloor \log_{10} (kn) \rfloor = 0$. Then we have $90$ numbers, namely $10$-$99$, for which $\lfloor \log_{10} (kn) \rfloor = 1$. The last number we have is $100$, which gives us $\lfloor \log_{10} (kn) \rfloor = 2$. This sum gives us only $90 + 2 = 92$, which is much too low. However, applying the same counting technique for $n = 10$, our sum comes out to be $9 + 180 + 3 = 192$, since there are $9$ terms for which $\lfloor \log_{10} (kn) \rfloor = 1$, $90$ terms for which $\lfloor \log_{10} (kn) \rfloor = 2$, and one term for which $\lfloor \log_{10} (kn) \rfloor = 3$. So we go up one more power of $10$ and get $18 + 270 + 4 = 292$, which is very close to what we are looking for.

Now we only have to bump up the value of $n$ a bit and check our sum. Each increase in $n$ by $1$ actually increases the value of our sum by $1$ as well (except for $n = 101$), because whenever a $4$ is added to the sum, a $3$ is taken away. It doesn't take long to check and see that the value of $n$ we're looking for is $\boxed{109}$ , which corresponds to a sum of exactly $300$.

~ anellipticcurveoverq