Help Please!

[matviy]

We have numbers 1,2,3...2003,2004,2005. In one move we take 2 random numbers and subtract them and replace them with the greatest value of the difference. eventually we have 1 number. Can that nunber be 0?


Pleaes help me. I have been banging my head aginst the wall for days and i still dont get it. THank you. :?

 

[tkhunny]

greatest value of the difference

What does that mean? Perhaps "absolute" value of the difference? In which case, the value never will be less than 1. It seems quite possible that the last two values could be 1 and 1, making the final value zero (0).

 

[soroban]

Hello, matviy!

I have a theory, but I can't follow through with it.

We have numbers 1,2,3...2003,2004,2005.
In one move we take 2 random numbers and subtract them
and replace them with the greatest value of the difference.
Eventually we have 1 number.
Can that number be 0?

Here's my theory . . . and if I'm totally wrong, someone please tell me!

To end up with a final 0, we want the differences to be as "equal" as possible.

One way is to make all the differences equal to 1.

Put the "1" aside . . . and subtract consecutive pairs of numbers.
. . |2 -3| = 1, |4 - 5| = 1, |6 - 7| = 1, ... , |2002 - 2003| = 1, |2004 - 2005| = 1

We have the initial "1" and 1002 more 1's.

With 1003 ones (an <u>odd</u> number),
. . I see no way to get a final difference of zero.


This makes me <u>suspect</u> that the answer is No.
. . But the proof eludes me.

 

[matviy]

Hello, matviy!

I have a theory, but I can't follow through with it.

We have numbers 1,2,3...2003,2004,2005.
In one move we take 2 random numbers and subtract them
and replace them with the greatest value of the difference.
Eventually we have 1 number.
Can that number be 0?

Here's my theory . . . and if I'm totally wrong, someone please tell me!

To end up with a final 0, we want the differences to be as "equal" as possible.

One way is to make all the differences equal to 1.

Put the "1" aside . . . and subtract consecutive pairs of numbers.
. . |2 -3| = 1, |4 - 5| = 1, |6 - 7| = 1, ... , |2002 - 2003| = 1, |2004 - 2005| = 1

We have the initial "1" and 1002 more 1's.

With 1003 ones (an <u>odd</u> number),
. . I see no way to get a final difference of zero.


This makes me <u>suspect</u> that the answer is No.
. . But the proof eludes me.

thank you. this mught work. But does any body have any other ideas too? please?

 

[Gene]

My stratagy is to start with simple layouts and look for a patern.
If n = 4x you can. Subtract x from n/2 +x to get an even number of the same numbers.
If n = 4x-1 you can. Subtract 1 from n getting two n-1s. subtract them getting n-3 numbers 2 thru n-2. (I assume you throw out zeros till the last.) Now the sticky part

  n   can
  1    N
  2    N
  3    Y
  4    Y
  5    N? Don't see how.
  6    N? Don't see how.
  7    Y
  8    Y

I suspect that's the patern. Still working on 4x+1 and 4x+2. I think its a parity thing.
2005 = 4x+1 (unfortunatly) so I think it can't go to zero.

 

[matviy]

hm...any morea ideas? Becasue nonne of these prove much. :?