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 Post subject: Numbers on a Blackboard
PostPosted: Fri, 6 Apr 2012 05:15:18 UTC 
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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:
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.


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

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