There are $2n$ consecutive integers on a board. It is permitted to split them into pairs and simultaneously replace each pair by their difference (not necessarily positive) and their sum. Prove that it is impossible to obtain any $2n$ consecutive integers again. Alexandr Gribalko
Problem
Source: Tournament of Towns, Senior A-Level Paper, Spring 2020 , p6
Tags: consecutive, number theory, combinatorics
Rizsgtp
03.06.2020 14:12
parmenides51 wrote: There are $2n$ consecutive integers on a board. It is permitted to split them into pairs and simultaneously replace each pair by their difference (not necessarily positive) and their sum. Prove that it is impossible to obtain any $2n$ consecutive integers again. Alexandr Gribalko Actually , you want to tell that we take , two consecutive integers and we replace them with their sum and difference this is the meaning right ??
parmenides51
03.06.2020 15:56
you do not replace 2 consecutive, you replace every pair of the different n pairs totally, by their difference and their sum, it doesn't have each pair to have consecutive numbers e.g. you take the numbers $5,6,7,8$ as pairs $5-8$ and $6-7$, replacing them we might get $-3,1, 13,13$ or $1, 3,13,13$ or $-3,-1, 13,13$ or $-1, -3,13,13$ you might gave taken them as pairs $5-6$ and $7-8$ or another case might be $5-7$ and $6-8$ PS. that is what I understand by this official wording
parmenides51
03.06.2020 16:23
in 2n consecutive we have n odd and n even numbers
sum / differrence of 2 odds = even,
sum / differrence of 2 evens = even,
sum / differrence of 1 even and 1odd = odd
so if we take any pair of the same parity, their sum and difference are 2 even numbers
while if we take any pair of the different parity, their sum and difference are 2 odd numbers
in order to get 2n consecutive at the end there is only 1 case, to have n even and n odd numbers
in order to have n even numbers , we have used n numbers of the same parity, either n odds (1) or n evens (2) (only one of these cases)
in order to have n odd numbers , we have used n numbers of different parity, which happens only in the case n/2 odd and n/2 even numbers (3)
so in any case we should have at the start (only one of these cases):
- case 1: from (1) , (3) we get 3n/2 odd and n/2 even numbers
- case 2: from (2) , (3) we get 3n/2 even and n/2 odd numbers
none of this case can happen since at the start we have n even and n odds
PS. please tell me whether it is correct
Pathological
03.06.2020 23:53
Consider the variable $S$ which is the sum of the squares of all numbers on the board. Show that:
$(1)$ $S$ doubles whenever we apply the operation.
$(2)$ $v_2(S) = v_2(n)$ whenever the numbers on the board are $2n$ consecutive integers.
Neither of the above is hard to show, and so we're done.