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

(Solution 3)
(Solution 3)
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== 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(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 < 100</math> (that is, a sum greater than <math>300</math>) and the sum is obviously monotone in <math>n</math>, the largest value is <math>\boxed{109}</math>.
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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(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>.
  
 
== See also ==
 
== See also ==

Revision as of 18:31, 21 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 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 $400 - M - N = 300$, from which $M + N = 100$. Now $M+1$ is the smallest integer $k$ for which $\log(kn) >= 4$ or $kn >= 10000$, thus \[M = \left\lceil\frac{10000}{n}\right\rceil - 1.\] Similarly, \[N = \left\lceil\frac{1000}{n}\right\rceil - 1.\] Unless $n$ is a divisor of $10000$, we have \[M = \left\lfloor\frac{10000}{n}\right\rfloor\] and \[N = \left\lfloor\frac{1000}{n}\right\rfloor = \left\lfloor \frac{M}{10} \right\rfloor.\] Under the same assumption, \[M + \left\lfloor \frac{M}{10} \right\rfloor = 100,\] and so $M = 91$. Dividing $10000$ by $91$ we see that $n = 109$. Since $n = 110$ is also not a divisor of $10000$ and with $n = 110$ we get $M + N = 90 + 9 = 99 < 100$ (thus a sum of $301 > 300$) and the sum is obviously monotone in $n$, the largest value is $\boxed{109}$.

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
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