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

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Therefore, <cmath>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.</cmath> Since we want the largest possible <math>n</math>, the answer is  <math>\boxed{109}</math>.
 
Therefore, <cmath>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.</cmath> Since we want the largest possible <math>n</math>, the answer is  <math>\boxed{109}</math>.
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=== Solution 4 ===
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Since we're working with base-<math>10</math> logarithms, we can start by testing out <math>n</math>'s that are powers of <math>10</math>. For <math>n = 1</math>, the terms in the sum are <math>\lfloor \log_{10} (1)\rfloor, \lfloor \log_{10} (2)\rfloor, \lfloor \log_{10} (3) \rfloor , . . . , \lfloor \log_{10} (100) \rfloor</math>. For numbers <math>1</math>-<math>9</math>, <math>\lfloor \log_{10} (kn) \rfloor = 0</math>. Then we have <math>90</math> numbers, namely <math>10</math>-<math>99</math>, for which <math>\lfloor \log_{10} (kn) \rfloor = 1</math>. The last number we have is <math>100</math>, which gives us <math>\lfloor \log_{10} (kn) \rfloor = 2</math>. This sum gives us only <math>90 + 2 = 92</math>, which is much too low. However, applying the same counting technique for <math>n = 10</math>, our sum comes out to be <math>9 + 180 + 3 = 192</math>, since there are <math>9</math> terms for which <math>\lfloor \log_{10} (kn) \rfloor = 1</math>, <math>90</math> terms for which <math>\lfloor \log_{10} (kn) \rfloor = 2</math>, and one term for which <math>\lfloor \log_{10} (kn) \rfloor = 3</math>. So we go up one more power of <math>10</math> and get <math>18 + 270 + 4 = 292</math>, which is very close to what we are looking for.
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Now we only have to bump up the value of <math>n</math> a bit and check our sum. Each increase in <math>n</math> by <math>1</math> actually increases the value of our sum by <math>1</math> as well, so it doesn't take long to check and see that the value of <math>n</math> we're looking for is <math>\boxed{109}</math> , which corresponds to a sum of exactly <math>300</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 17:23, 15 December 2019

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

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, so 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 $300$.

See Also

2010 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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