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 Post subject: Numbers on a BlackboardPosted: Fri, 6 Apr 2012 05:15:18 UTC
 S.O.S. Oldtimer

Joined: Wed, 4 Apr 2012 03:51:40 UTC
Posts: 173
Location: Hockeytown
Suppose we have positive integers on a blackboard.

In what is defined as a "sumop" we replace 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 and , then we would have:

22 22 9 6 1

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

The solution is kind of contrived.~

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

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

We show that all even works.
Let . Consider any -tuple . Form the -tuple where . Observe that we can perform a "sumop" on to get (what I mean is that we add ). So, we get all of the 's to have the same highest power of dividing each of them (like , , ). Let this new -tuple be

Let be the largest odd divisor of . Let and be positive integers such that: . We perform a "sumop" on and . Since we note that we have a monovariant. So, since the can't decrease infinitely, we find that eventually they are equal. Hence, we have all of them equal eventually.

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