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 can rearrange the equation into . Obviously, the RHS is an integer, so for some integer . We can therefore make the substitution to get
(We'll try the case where later.) Now let for a nonnegative integer , so that .
Going back to , this implies . Making the substitution gives the system of inequalities
The first inequality simplifies to , or . Since is even, we must have , so . Therefore, . The second inequality simplifies to , or . Therefore, as is even, we have , or . Hence or Since both inequalities must be satisfied, we see that only , , , and satisfy both inequalities. Each will lead to a distinct solution for , and thus for , for a total of positive solutions.
Now let . We have
Since , this can be rewritten as
Since is positive, the least possible value of is , hence , i.e., it must be negative. But it is also equal to , which is positive, and this is a contradiction. Therefore, there are no negative roots.
The total number of roots to this equation is thus , or .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.