- #1
arpitm08
- 50
- 0
Need help on proof!
Question:
Is it possible to arrange the numbers 1, 2,..., 2009 in a row so that each number, with the exception of the two end numbers, is either the sum or the absolute value of the difference of the two numbers to either side of it?
Attempt:
Let's assume that three numbers a,b,c are in order somewhere in the row of a possible combination. Then b=c-a or c+a. Then the next number could be d=c-b or c+b. For the two values of d we can substitute for b and get, c-(c+or-a)=a and c+(c+or-a)=2c+or-a, but since we can't repeat values, it can't be a. so we would have 2c+or-a as the next term. So far we have a,b,c,2c+or-a. Now to get the next term, we can do the same thing. We would get e (the fifth term)=c+or-a or 3c+or-a. This way is not getting any simple, and there are way too many terms to do this for.
Could someone help me out? Thank You in advance.
Question:
Is it possible to arrange the numbers 1, 2,..., 2009 in a row so that each number, with the exception of the two end numbers, is either the sum or the absolute value of the difference of the two numbers to either side of it?
Attempt:
Let's assume that three numbers a,b,c are in order somewhere in the row of a possible combination. Then b=c-a or c+a. Then the next number could be d=c-b or c+b. For the two values of d we can substitute for b and get, c-(c+or-a)=a and c+(c+or-a)=2c+or-a, but since we can't repeat values, it can't be a. so we would have 2c+or-a as the next term. So far we have a,b,c,2c+or-a. Now to get the next term, we can do the same thing. We would get e (the fifth term)=c+or-a or 3c+or-a. This way is not getting any simple, and there are way too many terms to do this for.
Could someone help me out? Thank You in advance.