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Unread 17 Feb 2007, 14:45   #1
KoeN
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a (simple, but fun) riddle

- you have 51 coins, 11 of them being 'heads up'.
- you now have to make 2 groups; both of them having the same amount of coins that have heads up.
- you are blindfolded, can't ask for help and can't sense in whatever way which coins are head up or not.
- you are only allowed to flip the coins whenever you want.

how do you do this?


goodluck team.





*hint 1: there is a proper solution for this, no lame answers etc.
*hint 2: the number of coins for this riddle is irrelevant.
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Unread 17 Feb 2007, 14:51   #2
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Re: a (simple, but fun) riddle

Have two groups with all the coins on their sides. Admittedly, that's a cheating answer, and may or may not be disallowed by the rules depending on your interpretation.
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Unread 17 Feb 2007, 15:04   #3
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Re: a (simple, but fun) riddle

Get rid of 49 coins and hope for the best with the final two?

Best I can do in two minutes I'm afraid.
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Unread 17 Feb 2007, 15:08   #4
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Re: a (simple, but fun) riddle

i'd expect you to do better than that jakiri.
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Unread 17 Feb 2007, 15:17   #5
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Re: a (simple, but fun) riddle

You turn over all the coins just once, and split the group randomly into two?
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Unread 17 Feb 2007, 15:17   #6
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Re: a (simple, but fun) riddle

I'm not sure how exactly it works but I'm hoping the solution involves beheading a daft dutchman who posts stupid uninteresting riddles on internet forums.
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Unread 17 Feb 2007, 15:26   #7
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Re: a (simple, but fun) riddle

make two groups of one coin, balanced on their edge. Neither have a coin which is heads up therefore both have the same number
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Unread 17 Feb 2007, 15:31   #8
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Re: a (simple, but fun) riddle

Alternately, take 11 coins and flip them. This makes pile one, and the rest make pile 2.

If a coin you pick up was heads and you flip it, then it is tails in pile 1 and pile 2 has one less heads.
If a coin you pick up was tails and you flip it, then it is heads in pile 1 and equals one of the other heads which are in pile 2
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Unread 17 Feb 2007, 15:44   #9
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Re: a (simple, but fun) riddle

worked example :

51 coins, 11 are heads = 40 tails

Take one coin, its heads. Flipped and is placed into pile 1.
1) T , 2) 10H,40T
Take one coin, its tails. Flipped and is placed into pile 1.
1) T,H , 2) 10H,39T
Take one coin, its tails. Flipped and is placed into pile 1.
1) T,2H , 2) 10H,38T
Take one coin, its heads. Flipped and is placed into pile 1.
1) 2T,2H, 2) 9H,38T.
Take one coin, its heads. Flipped and is placed into pile 1.
1) 3T,2H, 2) 8H,38T.
Take one coin, its heads. Flipped and is placed into pile 1.
1) 4T,2H, 2) 7H,38T.
Take one coin, its tails. Flipped and is placed into pile 1.
1) 4T,3H, 2) 7H, 37T.
Take one coin, its tails. Flipped and is placed into pile 1.
1) 4T,4H, 2) 7H, 36T.
Take one coin, its tails. Flipped and is placed into pile 1.
1) 4T,5H, 2) 7H, 35T.
Take one coin, its tails. Flipped and is placed into pile 1.
1) 4T,6H, 2) 7H, 34T
Take one coin, its heads. Flipped and is placed into pile 1.
1) 5T,6H, 2) 6H, 34T.

Both piles have 6 heads in them, fitting the requirements.
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Unread 17 Feb 2007, 15:47   #10
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Re: a (simple, but fun) riddle

it's much simpler phil.
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Unread 17 Feb 2007, 15:49   #11
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Re: a (simple, but fun) riddle

Elaborate then. My method works and i have shown as much. ( and if you go " oh both piles one and two are empty" as a cheat answer then may bodily harm be inflicted upon you )

btw was this an assignment of yours to do which has now been completed?
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Unread 17 Feb 2007, 16:04   #12
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Re: a (simple, but fun) riddle

Quote:
Originally Posted by KoeN
it's much simpler phil.
I think he's right, although it probably doesn't matter how you make the initial division.
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Unread 17 Feb 2007, 16:16   #13
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Re: a (simple, but fun) riddle

OK:

if you have 11 coins that are head up, then pick 11 random coins from the big pile of 51.

for example: let's say 2 out of 11 are head up. this means 9 head up coins are still left on the big pile.

turn around all coins of the 11 you picked, and you automatically have 2 groups of 9 coins on both piles.


easy eh?


edit: well done phil!

Last edited by KoeN; 17 Feb 2007 at 16:21.
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Unread 17 Feb 2007, 16:17   #14
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Re: a (simple, but fun) riddle

Words fail me.
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Unread 17 Feb 2007, 16:20   #15
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Re: a (simple, but fun) riddle

The title said it was a fun riddle...
I want my money back!
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Unread 17 Feb 2007, 16:26   #16
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Re: a (simple, but fun) riddle

Heres one for you koen ( and others ) . You take a standard chessboard (8x8 = 64 squares) and remove two squares, one at the top left and the other at the bottom right. essentially the diagonal ones are taken off.
Can the remaining squares be covered by 31 dominos where each domino takes up two squares?
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Unread 17 Feb 2007, 16:27   #17
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Re: a (simple, but fun) riddle

Quote:
Originally Posted by Phil^
Can the remaining squares be covered by 31 dominos where each domino takes up two squares?
Yes
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Unread 17 Feb 2007, 16:29   #18
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Re: a (simple, but fun) riddle

Quote:
Originally Posted by Phil^
Heres one for you koen ( and others ) . You take a standard chessboard (8x8 = 64 squares) and remove two squares, one at the top left and the other at the bottom right. essentially the diagonal ones are taken off.
Can the remaining squares be covered by 31 dominos where each domino takes up two squares?
No. You've removed two squares of the same colour, whereas a domino must cover a black one and a white one.
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Unread 17 Feb 2007, 16:29   #19
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Re: a (simple, but fun) riddle

Quote:
Originally Posted by JonnyBGood
Words fail me.
Hes dutch



I get the principle ie take 11 random coins into one pile and 40 random coins in another. If by chance you picked completely correctly theres 11 heads in one pile and no heads in the other, by flipping the pile with 11 heads both piles have no heads. But lets say you get 1 head in the '40 pile' and 1 tail in the '11 pile' by flipping the 11 pile you've now got one head and the others are tails.


I 'get' the solution but what would a generalised formula for the riddle be*?? the flipping over bit is confusing me, if you have N coins

N=n(T)+n(H)

Pile 1 = N-n(H)
Pile 2 = N-n(T)

Flipping over pile 2 would be 'what' in maths terms?

Nod?



*its more interesting



edit:


I think we're going to need matrices

N [Matrix] = [Scalar matrix ie numbers].[Unit matrix for 'state'] + [Scalar matrix].[Unit matrix]^-1


edit edit: it can't be a unit matrix as the reciprocal would be identical, lets call it a state matrix instead
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Unread 17 Feb 2007, 16:31   #20
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Re: a (simple, but fun) riddle

Quote:
Originally Posted by MrL_JaKiri
No. You've removed two squares of the same colour, whereas a domino must cover a black one and a white one.
winner
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Unread 17 Feb 2007, 16:31   #21
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Re: a (simple, but fun) riddle

God, I hate riddles.
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Unread 17 Feb 2007, 19:55   #22
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Re: a (simple, but fun) riddle

You are given 5 bags. There are 10 beads in each of the bags. In four of the bags, the beads each weigh 10 kilograms. In the remaining bag, each bead weighs only 9 kilograms. All the bags and beads look identical. You must find out which bag has the lighter beads. The problem is that all the bags look identical and all the beads look identical. You can use a scale, but it has to be a single-tray scale, not a two-tray balance scale. Also, you may use the scale only once. How can you find out which bag has the lighter beads?
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Unread 17 Feb 2007, 20:20   #23
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Re: a (simple, but fun) riddle

all the methods i can think of require at least two measurements unless you are allowed to move the beads between bags, in which case you can move one bag into another, ending up with two bags. Weighing one bag, if its weight is 200kg then its in the other bag. Otherwise its that one
ALso theres the put them all into one bag method of doing it.
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Unread 17 Feb 2007, 20:24   #24
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Re: a (simple, but fun) riddle

You can't mix the bags.
You are allowed to take the beads out of the bags.
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Unread 17 Feb 2007, 20:31   #25
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Re: a (simple, but fun) riddle

I have a gun with one bullet in it and both koen and alessio are trying to talk to me about how great holland is, how do I kill them both?







Note this isn't so much a riddle as advanced planning.
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Unread 17 Feb 2007, 20:34   #26
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Re: a (simple, but fun) riddle

hm. then its got to involve taking advantage of the fact that every bead in the 'lighter' bag is lighter and devising a joint measurement which somehow illuminates this while not allowing you to get confused with the others.
Not sure precisely yet but i'll work on it in a bit.
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Unread 17 Feb 2007, 20:35   #27
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Re: a (simple, but fun) riddle

Quote:
Originally Posted by JonnyBGood
I have a gun with one bullet in it and both koen and alessio are trying to talk to me about how great holland is, how do I kill them both?







Note this isn't so much a riddle as advanced planning.
Don't take the risk, shoot yourself.
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Unread 17 Feb 2007, 20:40   #28
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Re: a (simple, but fun) riddle

Quote:
Originally Posted by milo
Hes dutch



I get the principle ie take 11 random coins into one pile and 40 random coins in another. If by chance you picked completely correctly theres 11 heads in one pile and no heads in the other, by flipping the pile with 11 heads both piles have no heads. But lets say you get 1 head in the '40 pile' and 1 tail in the '11 pile' by flipping the 11 pile you've now got one head and the others are tails.


I 'get' the solution but what would a generalised formula for the riddle be*?? the flipping over bit is confusing me, if you have N coins

N=n(T)+n(H)

Pile 1 = N-n(H)
Pile 2 = N-n(T)

Flipping over pile 2 would be 'what' in maths terms?

Nod?
I'm not sure what youre asking.

If you just want a proof, then assume you have N coins of which H are heads. Split the coins into 2 piles by selecting H random coins (call this pile 1), and call the coins left over pile 2.

Let H1 be the number of heads in pile 1, and H2 be the number of heads in pile 2. Since H = H1 + H2, the number of heads left in pile 2 is H2 = H - H1. So we need to show there are H - H1 coins in pile 1.

pile 1 contains H coins, of which H1 are heads (by definition). Affter you flip them all over, you obviously have H - H1 heads. So both piles now have an equal number of heads.
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Unread 17 Feb 2007, 21:29   #29
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Re: a (simple, but fun) riddle

Quote:
Originally Posted by jt25man
You can't mix the bags.
You are allowed to take the beads out of the bags.
n beads from the nth bag. The rest is really obvious.
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Unread 17 Feb 2007, 21:46   #30
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Re: a (simple, but fun) riddle

jakiris right, one bean from first bag, two beans from second, three from third and so on, the final weight of those beans tells you which bag has the lighter ones.
ie the final weight SHOULD be 150 if all had 10kg weights in it, but since one doesnt then theres gonna be an offset
if the offset is 1, ie 149 then it was the first bag. If the offset is 2 ie 148kg then it was the second bag, and so on
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Unread 18 Feb 2007, 01:27   #31
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Re: a (simple, but fun) riddle

Quote:
Originally Posted by JonnyBGood
I have a gun with one bullet in it and both koen and alessio are trying to talk to me about how great holland is, how do I kill them both?
You shoot one in the head, and just beat to death the other with the handle.

Quote:
Originally Posted by MrL_JaKiri
n beads from the nth bag. The rest is really obvious.
Quote:
Originally Posted by Phil^
jakiris right, one bean from first bag, two beans from second, three from third and so on, the final weight of those beans tells you which bag has the lighter ones.
ie the final weight SHOULD be 150 if all had 10kg weights in it, but since one doesnt then theres gonna be an offset
if the offset is 1, ie 149 then it was the first bag. If the offset is 2 ie 148kg then it was the second bag, and so on
That is correct.
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Unread 18 Feb 2007, 02:07   #32
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Re: a (simple, but fun) riddle

Quote:
Originally Posted by Nodrog
I'm not sure what youre asking.

Neither am i basically i wanted a 'maths statement that encompassed the process' somewhere beyond a proof.
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Unread 19 Feb 2007, 18:47   #33
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Re: a (simple, but fun) riddle

no it won't take off!!!
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