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

(Solution 3)
 
(16 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 +
__TOC__
 
== Problem ==
 
== Problem ==
For each positive integer n, let <math>f(n) = \sum_{k = 1}^{100} \lfloor log_{10} (kn) \rfloor</math>. Find the largest value of n for which <math>f(n) \le 300</math>.
+
For each positive integer n, let <math>f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor</math>. Find the largest value of <math>n</math> for which <math>f(n) \le 300</math>.
  
 
'''Note:''' <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>.
 
'''Note:''' <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>.
  
== Solution 1 ==
+
==Solution==
 +
=== Solution 1 ===
 
Observe that <math>f</math> is strictly increasing in <math>n</math>. We realize that we need <math>100</math> terms to add up to around <math>300</math>, so we need some sequence of <math>2</math>s, <math>3</math>s, and then <math>4</math>s.
 
Observe that <math>f</math> is strictly increasing in <math>n</math>. We realize that we need <math>100</math> terms to add up to around <math>300</math>, so we need some sequence of <math>2</math>s, <math>3</math>s, and then <math>4</math>s.
  
 
It follows that <math>n \approx 100</math>. Manually checking shows that <math>f(109) = 300</math> and <math>f(110) > 300</math>. Thus, our answer is <math>\boxed{109}</math>.
 
It follows that <math>n \approx 100</math>. Manually checking shows that <math>f(109) = 300</math> and <math>f(110) > 300</math>. Thus, our answer is <math>\boxed{109}</math>.
  
== Solution 2 ==
+
=== Solution 2 ===
Because we want the value for which <math>f(n)=300</math>, the average value of the 100 terms of the sequence should be around <math>3</math>. For the value of <math>\lfloor log_{10} (kn) \rfloor</math> to be <math>3</math>, <math>1000 \le kn < 10000</math>. We want kn to be around the middle of that range, and for k to be in the middle of 0 and 100, let <math>k=50</math>, so <math>50n=\frac{10000+1000}{2}=\frac{11000}{2}=5500</math>, and <math>n = 110</math>. <math>f(110) = 301</math>, so we want to lower <math>n</math>. Testing <math>109</math> yields <math>300</math>, so our answer is still <math>\boxed{109}</math>.
+
Because we want the value for which <math>f(n)=300</math>, the average value of the 100 terms of the sequence should be around <math>3</math>. For the value of <math>\lfloor \log_{10} (kn) \rfloor</math> to be <math>3</math>, <math>1000 \le kn < 10000</math>. We want kn to be around the middle of that range, and for k to be in the middle of 0 and 100, let <math>k=50</math>, so <math>50n=\frac{10000+1000}{2}=\frac{11000}{2}=5500</math>, and <math>n = 110</math>. <math>f(110) = 301</math>, so we want to lower <math>n</math>. Testing <math>109</math> yields <math>300</math>, so our answer is still <math>\boxed{109}</math>.
  
== Solution 3 ==
+
=== Solution 3 ===
  
For any <math>n</math> where the sum is close to <math>300</math>, all the terms in the sum must be equal to <math>2</math>, <math>3</math> or <math>4</math>. Let <math>M</math> be the number of terms less than or equal to <math>3</math> and <math>N</math> be the number of terms equal to <math>2</math> (also counted in <math>M</math>). With this definition of <math>M</math> and <math>N</math> the total will be <math>400 - M - N = 300</math>, from which <math>M + N = 100</math>. Now <math>M+1</math> is the smallest integer <math>k</math> for which <math>\log_{10}(kn) >= 4</math> or <math>kn >= 10000</math>, thus <cmath>M = \left\lceil\frac{10000}{n}\right\rceil - 1.</cmath> Similarly, <cmath>N = \left\lceil\frac{1000}{n}\right\rceil - 1.</cmath> Unless <math>n</math> is a divisor of <math>10000</math>, we have <cmath>M = \left\lfloor\frac{10000}{n}\right\rfloor</cmath> and <cmath>N = \left\lfloor\frac{1000}{n}\right\rfloor = \left\lfloor \frac{M}{10} \right\rfloor.</cmath> Under the same assumption, <cmath>M + \left\lfloor \frac{M}{10} \right\rfloor = 100,</cmath> and so <math>M = 91</math>. Dividing <math>10000</math> by <math>91</math> we see that <math>n = 109</math>. Since <math>n = 110</math> is also not a divisor of <math>10000</math> and with <math>n = 110</math> we get <math>M + N = 90 + 9 = 99 < 100</math> (thus a sum of <math>301 > 300</math>) and the sum is obviously monotone in <math>n</math>, the largest value is <math>\boxed{109}</math>.
+
For any <math>n</math> where the sum is close to <math>300</math>, all the terms in the sum must be equal to <math>2</math>, <math>3</math> or <math>4</math>. Let <math>M</math> be the number of terms less than or equal to <math>3</math> and <math>N</math> be the number of terms equal to <math>2</math> (also counted in <math>M</math>). With this definition of <math>M</math> and <math>N</math> the total will be <math>400 - M - N \le 300</math>, from which <math>M + N \ge 100</math>. Now <math>M+1</math> is the smallest integer <math>k</math> for which <math>\log_{10}(kn) \ge 4</math> or <math>kn \ge 10000</math>, thus <cmath>M = \left\lfloor\frac{9999}{n}\right\rfloor.</cmath> Similarly, <cmath>N = \left\lfloor\frac{999}{n}\right\rfloor = \left\lfloor\frac{M}{10}\right\rfloor.</cmath>
  
== See also ==
+
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>.
 +
 
 +
=== Solution 4 (similar to Solution 1, but with more detail) ===
 +
 
 +
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.
 +
 
 +
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 (except for <math>n = 101</math>), because whenever a <math>4</math> is added to the sum, a <math>3</math> is taken away. 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>.
 +
 
 +
~ anellipticcurveoverq
 +
 
 +
==Video Soln==
 +
 
 +
https://youtu.be/K2xPdIti_AI
 +
 
 +
== See Also ==
 
{{AIME box|year=2010|num-b=13|num-a=15|n=I}}
 
{{AIME box|year=2010|num-b=13|num-a=15|n=I}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 18:20, 28 June 2020

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

Video Soln

https://youtu.be/K2xPdIti_AI

See Also

2010 AIME I (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS