Difference between revisions of "2007 AIME II Problems/Problem 7"
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== Problem == | == Problem == | ||
− | Given a real number <math>x,</math> let <math>\lfloor x \rfloor</math> denote the greatest integer less than or equal to <math>x.</math> For a certain integer <math>k,</math> there are exactly <math>70</math> positive integers <math>n_{1}, n_{2}, \ldots, n_{70}</math> such that <math>k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{1}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor</math> and <math>k</math> divides <math>n_{i}</math> for all <math>i</math> such that <math>1 \leq i \leq 70.</math> | + | Given a [[real number]] <math>x,</math> let <math>\lfloor x \rfloor</math> denote the [[floor function|greatest integer]] less than or equal to <math>x.</math> For a certain [[integer]] <math>k,</math> there are exactly <math>70</math> positive integers <math>n_{1}, n_{2}, \ldots, n_{70}</math> such that <math>k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{1}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor</math> and <math>k</math> divides <math>n_{i}</math> for all <math>i</math> such that <math>1 \leq i \leq 70.</math> |
Find the maximum value of <math>\frac{n_{i}}{k}</math> for <math>1\leq i \leq 70.</math> | Find the maximum value of <math>\frac{n_{i}}{k}</math> for <math>1\leq i \leq 70.</math> | ||
== Solution == | == Solution == | ||
− | For <math>x = 1</math>, we see that <math>\sqrt[3]{1} \ldots \sqrt[3]{7}</math> all work, giving 7 integers. For <math>x=2</math>, we see that in <math>\sqrt[3]{8} \ldots \sqrt[3]{26}</math>, all of the even numbers work, giving 10 integers. For <math>x = 3</math>, we get 13, and so on. We can predict that at <math>x = 22</math> we get 70. | + | For <math>x = 1</math>, we see that <math>\sqrt[3]{1} \ldots \sqrt[3]{7}</math> all work, giving 7 integers. For <math>x=2</math>, we see that in <math>\sqrt[3]{8} \ldots \sqrt[3]{26}</math>, all of the [[even]] numbers work, giving 10 integers. For <math>x = 3</math>, we get 13, and so on. We can predict that at <math>x = 22</math> we get 70. |
To prove this, note that all of the numbers from <math>\sqrt[3]{x^3} \ldots \sqrt[3]{(x+1)^3 - 1}</math> divisible by <math>x</math> work. Thus, <math>\frac{(x+1)^3 - 1 - x^3}{x} + 1 = \frac{3x^2 + 3x + 1 - 1}{x} + 1 = 3x + 4</math> (the one to be inclusive) integers will fit the conditions. <math>3k + 4 = 70 \Longrightarrow k = 22</math>. | To prove this, note that all of the numbers from <math>\sqrt[3]{x^3} \ldots \sqrt[3]{(x+1)^3 - 1}</math> divisible by <math>x</math> work. Thus, <math>\frac{(x+1)^3 - 1 - x^3}{x} + 1 = \frac{3x^2 + 3x + 1 - 1}{x} + 1 = 3x + 4</math> (the one to be inclusive) integers will fit the conditions. <math>3k + 4 = 70 \Longrightarrow k = 22</math>. |
Revision as of 19:50, 29 March 2007
Problem
Given a real number let denote the greatest integer less than or equal to For a certain integer there are exactly positive integers such that and divides for all such that
Find the maximum value of for
Solution
For , we see that all work, giving 7 integers. For , we see that in , all of the even numbers work, giving 10 integers. For , we get 13, and so on. We can predict that at we get 70.
To prove this, note that all of the numbers from divisible by work. Thus, (the one to be inclusive) integers will fit the conditions. .
The maximum value of . Therefore, the solution is .
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |