2007 AIME II Problems/Problem 7
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
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 .
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 !)
Video Solution by the Beauty of Math
|2007 AIME II (Problems • Answer Key • Resources)|
|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.