2007 AIME II Problems/Problem 7
From AoPSWiki
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 ![k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{1}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/6/9/0/69053ef8a8039840a7186514c5ef9697e1848894.gif)
and 
divides 
for all 
such that 
Find the maximum value of 
for
Solution
For 
, we see that ![\sqrt[3]{1} \ldots \sqrt[3]{7}](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/c/6/4/c64a568bb4918513bdeb1c1d1c557c99c8c838da.gif)
all work, giving 7 integers. For 
, we see that in ![\sqrt[3]{8} \ldots \sqrt[3]{26}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/2/a/e/2ae41ff8df758b501dd055f90f24cef1faf8e27b.gif)
, 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 ![\sqrt[3]{x^3} \ldots \sqrt[3]{(x+1)^3 - 1}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/8/f/d/8fdde096d872489b2814dbf19a02128466e6ac4f.gif)
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 • Resources) | ||
| Preceded by Problem 6 | Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
