|
17 Feb 2007, 14:45
|
#1
|
Lucky
Join Date: Jul 2003
Location: -
Posts: 3,830
|
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.
|
|
|
17 Feb 2007, 14:51
|
#2
|
The Twilight of the Gods
Join Date: Jan 2001
Posts: 23,481
|
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.
|
|
|
17 Feb 2007, 15:04
|
#3
|
I dunno...
Join Date: Jun 2003
Location: manchester
Posts: 1,502
|
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.
__________________
He shall drink naught but brine, for I'll not show him / Where the quick freshes are.
|
|
|
17 Feb 2007, 15:08
|
#4
|
Lucky
Join Date: Jul 2003
Location: -
Posts: 3,830
|
Re: a (simple, but fun) riddle
i'd expect you to do better than that jakiri.
|
|
|
17 Feb 2007, 15:17
|
#5
|
Not Dark or Handsome
Join Date: Nov 2001
Location: Cwmbru
Posts: 2,588
|
Re: a (simple, but fun) riddle
You turn over all the coins just once, and split the group randomly into two?
__________________
"You can't drink a pint of Bovril."
|
|
|
17 Feb 2007, 15:17
|
#6
|
Banned
Join Date: May 2001
Location: Further to the right
Posts: 19,441
|
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.
__________________
Some might ask what good is life without purpose but I'm anticipating a good lunch.
|
|
|
17 Feb 2007, 15:26
|
#7
|
Insomniac
Join Date: May 2003
Posts: 3,583
|
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
|
|
|
17 Feb 2007, 15:31
|
#8
|
Insomniac
Join Date: May 2003
Posts: 3,583
|
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
|
|
|
17 Feb 2007, 15:44
|
#9
|
Insomniac
Join Date: May 2003
Posts: 3,583
|
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.
|
|
|
17 Feb 2007, 15:47
|
#10
|
Lucky
Join Date: Jul 2003
Location: -
Posts: 3,830
|
Re: a (simple, but fun) riddle
it's much simpler phil.
|
|
|
17 Feb 2007, 15:49
|
#11
|
Insomniac
Join Date: May 2003
Posts: 3,583
|
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?
|
|
|
17 Feb 2007, 16:04
|
#12
|
Clerk
Join Date: Jun 2001
Posts: 13,940
|
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.
|
|
|
17 Feb 2007, 16:16
|
#13
|
Lucky
Join Date: Jul 2003
Location: -
Posts: 3,830
|
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.
|
|
|
17 Feb 2007, 16:17
|
#14
|
Banned
Join Date: May 2001
Location: Further to the right
Posts: 19,441
|
Re: a (simple, but fun) riddle
Words fail me.
__________________
Some might ask what good is life without purpose but I'm anticipating a good lunch.
|
|
|
17 Feb 2007, 16:20
|
#15
|
Rawr rawr
Join Date: Dec 2000
Location: Upside down
Posts: 5,300
|
Re: a (simple, but fun) riddle
The title said it was a fun riddle...
I want my money back!
__________________
"Yay"
|
|
|
17 Feb 2007, 16:26
|
#16
|
Insomniac
Join Date: May 2003
Posts: 3,583
|
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?
|
|
|
17 Feb 2007, 16:27
|
#17
|
Rawr rawr
Join Date: Dec 2000
Location: Upside down
Posts: 5,300
|
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
__________________
"Yay"
|
|
|
17 Feb 2007, 16:29
|
#18
|
The Twilight of the Gods
Join Date: Jan 2001
Posts: 23,481
|
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.
|
|
|
17 Feb 2007, 16:29
|
#19
|
Registered User
Join Date: Feb 2006
Posts: 1,094
|
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
Last edited by milo; 17 Feb 2007 at 17:19.
|
|
|
17 Feb 2007, 16:31
|
#20
|
Insomniac
Join Date: May 2003
Posts: 3,583
|
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
|
|
|
17 Feb 2007, 16:31
|
#21
|
Rawr rawr
Join Date: Dec 2000
Location: Upside down
Posts: 5,300
|
Re: a (simple, but fun) riddle
God, I hate riddles.
__________________
"Yay"
|
|
|
17 Feb 2007, 19:55
|
#22
|
Victim of Marriage
Join Date: Oct 2005
Location: NW Indiana
Posts: 784
|
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?
__________________
You mean there's life outside the internet...oh man I'm screwed.
|
|
|
17 Feb 2007, 20:20
|
#23
|
Insomniac
Join Date: May 2003
Posts: 3,583
|
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.
|
|
|
17 Feb 2007, 20:24
|
#24
|
Victim of Marriage
Join Date: Oct 2005
Location: NW Indiana
Posts: 784
|
Re: a (simple, but fun) riddle
You can't mix the bags.
You are allowed to take the beads out of the bags.
__________________
You mean there's life outside the internet...oh man I'm screwed.
|
|
|
17 Feb 2007, 20:31
|
#25
|
Banned
Join Date: May 2001
Location: Further to the right
Posts: 19,441
|
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.
__________________
Some might ask what good is life without purpose but I'm anticipating a good lunch.
|
|
|
17 Feb 2007, 20:34
|
#26
|
Insomniac
Join Date: May 2003
Posts: 3,583
|
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.
|
|
|
17 Feb 2007, 20:35
|
#27
|
Registered User
Join Date: Jan 2005
Location: London
Posts: 3,347
|
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.
__________________
The 20th century has been characterised by three developments of great political importance. The growth of democracy; the growth of corporate power; and the growth of corporate propaganda as a means of protecting corporate power against democracy.
|
|
|
17 Feb 2007, 20:40
|
#28
|
Registered User
Join Date: Jun 2000
Posts: 8,476
|
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.
|
|
|
17 Feb 2007, 21:29
|
#29
|
The Twilight of the Gods
Join Date: Jan 2001
Posts: 23,481
|
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.
|
|
|
17 Feb 2007, 21:46
|
#30
|
Insomniac
Join Date: May 2003
Posts: 3,583
|
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
|
|
|
18 Feb 2007, 01:27
|
#31
|
Victim of Marriage
Join Date: Oct 2005
Location: NW Indiana
Posts: 784
|
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.
__________________
You mean there's life outside the internet...oh man I'm screwed.
|
|
|
18 Feb 2007, 02:07
|
#32
|
Registered User
Join Date: Feb 2006
Posts: 1,094
|
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.
|
|
|
19 Feb 2007, 18:47
|
#33
|
:alpha:
Join Date: May 2002
Location: London, UK
Posts: 7,871
|
Re: a (simple, but fun) riddle
no it won't take off!!!
__________________
"There is no I in team, but there are two in anal fisting"
|
|
|
|
All times are GMT +1. The time now is 03:54.
| |