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 [[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_{ | + | 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_{2}}\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> |
Revision as of 06:12, 25 February 2012
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 |