Problem:
How many positive integers less than are there such that the equation has a solution for ? (The notation denotes the greatest integer that is less than or equal to .)
Solution:
For a positive integer , consider the problem of counting the number of integers such that has a solution with . Then , and because , it follows that . Thus there are possible integer values of for which the equation has a solution. Because and , the desired number of values of is .
The problems on this page are the property of the MAA's American Mathematics Competitions