2003 OIM Problems/Problem 1
Problem
a) There are two sequences, each of 2003 consecutive integers, and a board with 2 rows and 2003 columns
Decide if it is always possible to distribute the numbers of the first sequence in the first row and those of the second sequence in the second row, in such a way that the results obtained by adding the two numbers in each column form a new sequence of 2003 consecutive numbers.
b) What if 2003 is replaced by 2004?
In both a) and b), if the answer is affirmative, explain how you would distribute the numbers, and if the answer is negative, justify why.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
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We have,
Sequence 1 (First Row):
Sequence 2 (Second Row):
We need to arrange these numbers such that the sum in each column forms a sequence of 2003 consecutive numbers.
Let
Suppose we directly distribute the numbers into each row:
First Row:
Second Row:
Then the sum in each column is:
We want
Suppose
For
This implies:
This is impossible for all
Therefore, we cannot distribute the numbers in such a way that the sums in each column form a new sequence of 2003 consecutive numbers.
The same analysis applies to 2004 integers:
Sequence 1 (First Row):
Sequence 2 (Second Row):
Following similar steps, the sum in each column
Again, it is impossible to form a new sequence of 2004 consecutive numbers.
For both 2003 and 2004 consecutive integers, it is impossible to distribute the numbers in the sequences into two rows in such a way that the sums of the numbers in each column form a new sequence of consecutive numbers.
Thus, the answer is:
For both 2003 and 2004 integers, it is not possible.
By AnYu