1996 USAMO Problems/Problem 6
Problem
Determine (with proof) whether there is a subset of the integers with the following property: for any integer there is exactly one solution of with .
Solution
Start with an incomplete subset , such that for any integer n, there is exactly zero or one solutions to with . Let be the smallest integer such that for any , . Note that for any and
Suppose is the smallest non-negative integer without a solution in yet. Clearly, . Generate and such that , and . Thus, we now have the solution .
Note: The values 10 and 5 can be replaced by any sufficiently large values such that the first is twice the second.
Now, we must prove that the addition of these two terms to does not result in an integer n that has two solutions. Of course, which previously had no solutions. Furthermore, .
- For any , , and .
- Since , we get that and
- Similarly, , and .
- Since , we get that and .
Since all of these sums (other than ) are either greater than or less than , they are all sums that previously had no solutions. Furthermore, none of these sums are duplicated, as sums of different forms are contained in disjoint ranges of integers.
Thus, we have proved that we can generate a subset such that all non-negative integers n have a unique solution .
For negative integers M that have no solutions in a similar proof holds, but instead generating the terms and .
For any integer M that currently has no solution in S, we can always add two terms and such that that do not result in duplicated sums.
Thus, there does exist a subset of the integers such that for any integer there is exactly one solution to with .
See Also
1996 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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