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The whole, natural numbers 1 through 50 are written on a board. The following operation is performed 49 times: select any two of the numbers on the board and write the absolute value of their difference on the board. Then erase the two original numbers you chose (therefore only the absolute value of the difference is left on the board in addition to the other natural numbers and differences that have already been on the board). Determine all possible values of the remaining number.


marked as duplicate by athin, Rubio May 9 '18 at 1:06

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  • 1
    $\begingroup$ Are you talking about two consecutive numbers, or any two numbers (e.g. 20 and 35)? $\endgroup$ – gparyani May 8 '18 at 18:23
  • $\begingroup$ Any two numbers (i.e. 20 and 35) sorry for the confusion! $\endgroup$ – user18842sos May 8 '18 at 20:47

Just working through it in a very rudimentary manner:

All odd numbers 1-49 are possible
Haven't found a way to arrive at 0 or an even number

Edit - Terrible explanation of said rudimentary manner:

Looking for smallest number

I took each consecutive pair (2-1, 4-3, etc.) which left 25 1s. Those became 12 0s and a single 1. The obvious end result is '1'.

Looking for largest number

The largest difference possible on the board would be 50-0=50, but there's no way I found to achieve that without another single non-zero number.
So, I took '1', '50', and each consecutive pair in between. That resulted in a 49 (50 - 1) and 24 1s, then the 49 and 12 0s, resulting in 49.

Poking around in between

From then, I just picked a couple different numbers, leaving consecutive pairs between them, which resulted in other odd numbers (1,8,pairs = 7; 1,42,pairs = 41; etc.).

Explanation for exclusion of certain numbers

I can't really explain my terrible 'methods' of trying to find a 0 or even result. I mostly just liked the question and wanted it to get some attention, then learn all the obvious things I overlooked (like @ManyPinkHats pointing out that the sum is odd, so even results are impossible).

  • 3
    $\begingroup$ There's definitely no way to arrive at any ROT13(rira ahzore). A quick way to see this is the invariant that ROT13(gur fhz bs gur ahzoref ba gur obneq vf nyjnlf bqq). $\endgroup$ – ManyPinkHats May 8 '18 at 16:41
  • $\begingroup$ How are you choosing to work through it? By starting with the 50 numbers or with a smaller subset? Thank you for your contributions! :) @hagfy $\endgroup$ – user18842sos May 8 '18 at 20:49

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