2005 Indonesia MO Problems/Problem 5
Contents
[hide]Problem
For an arbitrary real number , denotes the greatest integer not exceeding . Prove that there is exactly one integer which satisfy .
Solution
First we will show that is a solution. Then we will show that there are no other integers that are a solution.
To find the solution, note that the absolute value of is usually smaller than , so should be close to . Plugging in results in . Trying out results in , so that value is a solution.
To show that there are no other integers that are a solution, we will consider two cases: one where is less than and one where is greater than .
Case 1:
The highest integer that meets the conditions is , and in that condition, .
Assume that . Let , where are integers and is less than .
If , then . Additionally,
If , then . Additionally,
For both cases, , so by induction, there are no integers less than that satisfy the equation.
Case 2:
The lowest integer that meets the conditions is , and in that condition, .
Assume that . Once again, let , where are integers and is less than .
If , then . Additionally,
If , then . Additionally,
For both cases, , so by induction, there are no integers greater than that satisfy the equation.
Therefore, there is only one integer that satisfies the original equation.
Solution 2
Using Quotient-Remainder Theorem, let , where , , are integers and
Since , .
This simplifies down to
If , then , which violates the restriction that . If , then , which again violates the restriction that .
Thus, the only possible value for q is 1, and .
Solving for gives as our only answer
See Also
2005 Indonesia MO (Problems) | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | Followed by Problem 6 |
All Indonesia MO Problems and Solutions |