Difference between revisions of "1991 IMO Problems/Problem 6"
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For any two distinct non-negative integers <math>i \neq j</math>, represent them as base-<math>n</math> positional notation: | For any two distinct non-negative integers <math>i \neq j</math>, represent them as base-<math>n</math> positional notation: | ||
− | <math>i = i_m n^m + i_{m-1} n^{m-1} + \cdots + i_2 n^2 + i_1 n + i_0 | + | <math>i = i_m n^m + i_{m-1} n^{m-1} + \cdots + i_2 n^2 + i_1 n + i_0</math> |
+ | |||
+ | <math>j = j_m n^m + j_{m-1} n^{m-1} + \cdots + j_2 n^2 + j_1 n + j_0</math> | ||
where <math>i_m, i_{m-1}, \cdots, i_2, i_1, i_0, j_m, j_{m-1}, \cdots, j_2, j_1, j_0 \in \left\{ 0, 1, 2, \cdots, n - 1 \right\}</math> | where <math>i_m, i_{m-1}, \cdots, i_2, i_1, i_0, j_m, j_{m-1}, \cdots, j_2, j_1, j_0 \in \left\{ 0, 1, 2, \cdots, n - 1 \right\}</math> |
Revision as of 05:07, 23 January 2024
Contents
[hide]Problem
An infinite sequence of real numbers is said to be bounded if there is a constant such that for every .
Given any real number , construct a bounded infinite sequence such that for every pair of distinct nonnegative integers .
Solution 1
Since , the series is convergent; let be the sum of this convergent series. Let be the interval (or any bounded subset of measure ).
Suppose that we have chosen points satisfying
for all distinct . We show that we can choose such that holds for all distinct . The only new cases are when one number (WLOG ) is equal to , so we must guarantee that for all .
Let be the interval , of length . The points that are valid choices for are precisely the points of , so we must show that this set is nonempty. The total length is at most the sum of the lengths . This is .
Therefore the total measure of is , so has positive measure and thus is nonempty. Choosing any and continuing by induction constructs the desired sequence.
Solution 2
The argument above would not work for , since only converges for . But Osmun Nal argues in this video that satisfies the stronger inequality for all distinct ; in other words, this sequence simultaneously solves the problem for all simultaneously.
Solution 3
Let , then we have
Thus
For any non-negative integer , represent it as base- positional notation:
, where
We directly construct
We allow leading zeros for representing due to the leading zeros do not effect the value of
Then
Thus , which is bounded
For any two distinct non-negative integers , represent them as base- positional notation:
where
Let be the minimum number such that , that is,
Then we have
Without loss of generality, assume that , then we have
Thus we have
~Joseph Tsai, mgtsai@gmail.com
See Also
1991 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
All IMO Problems and Solutions |