S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Thu, 24 Apr 2014 14:58:38 UTC

All times are UTC [ DST ]




Post new topic Reply to topic  [ 1 post ] 
Author Message
 Post subject: Numbers on a Blackboard
PostPosted: Fri, 6 Apr 2012 05:15:18 UTC 
Offline
Senior Member
User avatar

Joined: Wed, 4 Apr 2012 03:51:40 UTC
Posts: 129
Location: Hockeytown aka Detroit
Suppose we have n \ge 2 positive integers on a blackboard.

In what is defined as a "sumop" we replace 2 of these numbers with their sum.

For example, if we had the numbers

17 5 9 6 1

on the blackboard and we performed a "sumop" on 17 and 5, then we would have:

22 22 9 6 1

Find all n such that after a finite number of "sumops" we can get n identical numbers given any n-tuple of positive integers.

The solution is kind of contrived.~

Solution Sketch:
Spoiler:
Let any such n satisfying the desired result be called good and let all other n be called bad. I claim that n is good iff n is even.

First, we show n odd is bad.
Assume for the sake of contradiction that there is an odd number n that is good. Let n=2m-1.
Consider the n-tuple: (2,2,......,2,1), where there are 2m-2 two's and 1 one. Assuming that this does indeed become an n tuple of equal positive integers, let this n-tuple be (j,j,j,.....,j,j). Note that j>2
Now, considering the reverse operation of the "sumop" or the "pomus"
in which (p,p) \rightarrow (q,p-q) we note that each pomus decreases both numbers. But, this will act on the numbers in pairs. As a result, one of the j's won't decrease.

Contradiction.

We show that all n even works.
Let n=2m. Consider any 2m-tuple (x_1,x_2,......,x_{2m}). Form the m-tuple (y_1,y_2,...,y_m) where y_i=(x_{2i-1}+x_{2i},x_{2i-1}+x_{2i}). Observe that we can perform a "sumop" on y_i to get 2y_i (what I mean is that we add (x_{2i-1}+x_{2i})+(x_{2i-1}+x_{2i})). So, we get all of the y_i's to have the same highest power of 2 dividing each of them (like y_1=12, y_2=36, y_3=28). Let this new m -tuple be (z_1,z_2,....,z_m).

Let f(z_i) be the largest odd divisor of z_i. Let r and s be positive integers such that: f(z_r) \leq f(z_i) \leq f(z_s). We perform a "sumop" on z_r and z_s. Since f(z_r+z_s)<f(z_s) we note that we have a monovariant. So, since the f(z_s)-f(z_r) can't decrease infinitely, we find that eventually they are equal. Hence, we have all of them equal eventually.

_________________
math puns are the first sine of madness
-JDR
:mrgreen:


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 1 post ] 

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA