Difference between revisions of "2010 AIME I Problems/Problem 14"
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'''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 == | + | == 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 == | ||
+ | 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>. | ||
== See also == | == See also == |
Revision as of 16:16, 20 March 2010
Contents
Problem
For each positive integer n, let . Find the largest value of n for which .
Note: is the greatest integer less than or equal to .
Solution 1
Observe that is strictly increasing in . We realize that we need terms to add up to around , so we need some sequence of s, s, and then s.
It follows that . Manually checking shows that and . Thus, our answer is .
Solution 2
Because we want the value for which , the average value of the 100 terms of the sequence should be around . For the value of to be , . We want kn to be around the middle of that range, and for k to be in the middle of 0 and 100, let , so , and . , so we want to lower . Testing yields , so our answer is still .
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 |