2007 AIME II Problems/Problem 7
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 ![$k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{2}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor$](http://latex.artofproblemsolving.com/a/2/a/a2a31bac0a2bc6d9943fbdf0ef0b1a3bc933895e.png)
and 
divides 
for all 
such that 
Find the maximum value of 
for 
Solution
Solution 1
For 
, we see that ![$\sqrt[3]{1} \ldots \sqrt[3]{7}$](http://latex.artofproblemsolving.com/9/3/3/9339e15771e9f6fd053bad1f11137bcb1e576985.png)
all work, giving 7 integers. For 
, we see that in ![$\sqrt[3]{8} \ldots \sqrt[3]{26}$](http://latex.artofproblemsolving.com/0/b/1/0b1b2a0bb4dfe13cdca56c74fc2872c55cc07fc7.png)
, 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://latex.artofproblemsolving.com/d/d/4/dd4c7646f276f40c0128e3ad6cdc26603f11629c.png)
divisible by 
work. Thus, 
(the one to be inclusive) integers will fit the conditions. 
.
The maximum value of 
. Therefore, the solution is 
.
Solution 2
Obviously is positive. Then, we can let 
equal 
and similarly let 
equal 
.
The wording of this problem (which uses "exactly") tells us that 
. Taking away from our inequality results in 
. Since 
, 
, and 
are all integers, this inequality is equivalent to 
. Since is positive, we can divide the inequality by to get 
. Clearly the only that satisfies is 
.
Then, 
is the maximum value of 
. (Remember we set equal to !)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
