1985 AHSME Problems/Problem 30
Problem
Let be the greatest integer less than or equal to . Then the number of real solutions to is
Solution
We rearrange the equation as , where the right-hand side is now clearly an integer, meaning that for some non-negative integer . Therefore, in the case where , substituting gives To proceed, let be the unique non-negative integer such that , so that and our equation reduces to
The above inequalities therefore become where the first inequality can now be rewritten as , i.e. . Since is even for all integers , we must in fact have The second inequality similarly simplifies to , i.e. . As is even, this is equivalent to so the values of satisfying both inequalities are , , , and . Since , each of these distinct values of gives a distinct solution for , and thus for , giving a total of solutions in the case.
As is already the largest of the answer choices, this suffices to show that the answer is , but for completeness, we will show that the case indeed gives no other solutions. If (and so ), we require and recalling that for all , this equation can be rewritten as Since is positive, the least possible value of is , but this means which is a contradiction. Therefore the case indeed gives no further solutions, confirming that the total number of solutions is precisely .
See Also
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