January 19th, 2018, 10:54 AM  #1 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,904 Thanks: 883  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? Last edited by skipjack; January 19th, 2018 at 07:21 PM. 
January 19th, 2018, 11:02 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,842 Thanks: 1068 Math Focus: Elementary mathematics and beyond 
$\frac{n(n+1)}{2}$ 
January 19th, 2018, 11:15 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,904 Thanks: 883  
January 19th, 2018, 11:20 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,842 Thanks: 1068 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, 11:28 AM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,904 Thanks: 883 
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, 11:45 AM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,842 Thanks: 1068 Math Focus: Elementary mathematics and beyond 
So it depends on the outcome of the cardflipping 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 11:58 AM. 
January 20th, 2018, 05:54 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,904 Thanks: 883  
January 20th, 2018, 06:11 AM  #8 
Senior Member Joined: Sep 2016 From: USA Posts: 415 Thanks: 229 Math Focus: Dynamical systems, analytic function theory, numerics  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, 06:29 AM  #9 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,904 Thanks: 883 
Thanks SDK. So the "aim" is to end up with all cards facing down after the (n1)th pick, right? 
January 20th, 2018, 12:09 PM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,904 Thanks: 883 
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! 