My Math Forum Unclear (to me) problem
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 January 19th, 2018, 11:54 AM #1 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,658 Thanks: 740 Unclear (to me) problem Found this: n cards are laid on a table, all face down. The 1st person selects 1 card at random and turns it over, the 2nd person selects 2 cards at random and turns them over, the 3rd person selects 3 cards at random and turns them over, and so on, until the nth person turns all the n cards over. Determine all n such that it is possible for all n cards to be face up at the end. Whadda heck does that mean? Thanks from greg1313 and Joppy Last edited by skipjack; January 19th, 2018 at 08:21 PM.
 January 19th, 2018, 12:02 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,733 Thanks: 996 Math Focus: Elementary mathematics and beyond $\frac{n(n+1)}{2}$
 January 19th, 2018, 12:15 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,658 Thanks: 740 Thought so too...but not so... S.O.S. Mathematics CyberBoard • View topic - Cards on a table
 January 19th, 2018, 12:20 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,733 Thanks: 996 Math Focus: Elementary mathematics and beyond Maybe the person who thinks they have the correct answer doesn't have the correct answer. It seems pretty straightforward to me - if the solution is something else then the problem is semantics, not math.
 January 19th, 2018, 12:28 PM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,658 Thanks: 740 Did you read the full thread? What is confusing: if you go first and pick a card (say #3) , you turn it over. I go next: if I pick #2 and #3, I turn both over: so #3 is now face down!!
 January 19th, 2018, 12:45 PM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,733 Thanks: 996 Math Focus: Elementary mathematics and beyond So it depends on the outcome of the card-flipping and thus can be any $n$. Not much of a problem, IMO - but then I'm getting snappy in my old age. Consider T = top, B = bottom. B B B -> B B T -> B B B -> T T T I think it depends on the selection of cards but in theory (mine) any $n$ would work. Actually, 2 won't work. Yes, I get it now. It's actually more interesting than I originally thought. Last edited by greg1313; January 19th, 2018 at 12:58 PM.
January 20th, 2018, 06:54 AM   #7
Math Team

Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 11,658
Thanks: 740

Quote:
 Originally Posted by greg1313 Yes, I get it now. It's actually more interesting than I originally thought.
Seems to me that problem would be same if
the cards were not picked at random.

Can you give an example of 1 case where this is possible?
I simply can't. Except for n=1, of course!

January 20th, 2018, 07:11 AM   #8
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Joined: Sep 2016
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Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by Denis Seems to me that problem would be same if the cards were not picked at random. Can you give an example of 1 case where this is possible? I simply can't. Except for n=1, of course!
You are correct. Since the card flipping is random but they only ask for $n$ such that winning'' is possible, then the problem is equivalent if you allow the players to methodically pick cards.

 January 20th, 2018, 07:29 AM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,658 Thanks: 740 Thanks SDK. So the "aim" is to end up with all cards facing down after the (n-1)th pick, right?
 January 20th, 2018, 01:09 PM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,658 Thanks: 740 Think I finally got this damn thing! d = down, u = up - 1 2 3 4=n 0 d d d d 1 u d d d 2 u u u d 3 d d d d 4 u u u u Solution seems to be n@4 = 0 No idea how to prove it!

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