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January 19th, 2018, 10:54 AM   #1
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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?
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Last edited by skipjack; January 19th, 2018 at 07:21 PM.
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January 19th, 2018, 11:02 AM   #2
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$\frac{n(n+1)}{2}$

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January 19th, 2018, 11:15 AM   #3
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Thought so too...but not so...

S.O.S. Mathematics CyberBoard • View topic - Cards on a table
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January 19th, 2018, 11:20 AM   #4
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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.
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January 19th, 2018, 11:28 AM   #5
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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!!
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January 19th, 2018, 11:45 AM   #6
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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 11:58 AM.
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January 20th, 2018, 05:54 AM   #7
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Quote:
Originally Posted by greg1313 View Post
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!
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January 20th, 2018, 06:11 AM   #8
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Quote:
Originally Posted by Denis View Post
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.
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January 20th, 2018, 06:29 AM   #9
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Thanks SDK.

So the "aim" is to end up with all cards facing down
after the (n-1)th pick, right?
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January 20th, 2018, 12:09 PM   #10
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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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