A square $ n \times n $, ($ n> 2 $) contains nonzero real numbers. It is known that every number is exactly $ k $ times smaller than the sum of all the numbers in its row or sum of all number in its column. For which real numbers $ k $ is this possible?
Problem
Source: 239 2001 J1
Tags: combinatorics
20.05.2020 22:30
The correct translation is: A square $n \times n$, ($n>2$) contains nonzero numbers. It is known that each number is exactly $k$ times less than the sum of the numbers in its cross (i.e. the numbers in its row and column without the number itself). For which $k$ is this possible?
21.05.2020 22:11
nikabarakovich wrote: The correct translation is: A square $n \times n$, ($n>2$) contains nonzero numbers. It is known that each number is exactly $k$ times less than the sum of the numbers in its cross (i.e. the numbers in its row and column without the number itself). For which $k$ is this possible? Do you mean the numbers are real?
21.05.2020 22:39
Surely we can just take the square: r r r r ... r r r r r ... r .... r r r r ... r and then we require: $$2(n-1)r - k = r \Rightarrow k = 2(n-2)r$$and for $n > 2$ this function is surjective so all $k$ work. I may have horribly misinterpreted the question..
22.05.2020 09:14
matinyousefi wrote: nikabarakovich wrote: The correct translation is: A square $n \times n$, ($n>2$) contains nonzero numbers. It is known that each number is exactly $k$ times less than the sum of the numbers in its cross (i.e. the numbers in its row and column without the number itself). For which $k$ is this possible? Do you mean the numbers are real? Yes, the word числа means numbers.