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.

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.

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.

Re: a (simple, but fun) riddle
 

i'd expect you to do better than that jakiri. :(

Re: a (simple, but fun) riddle
 

You turn over all the coins just once, and split the group randomly into two?

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.

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

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

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.

Re: a (simple, but fun) riddle
 

it's much simpler phil. :)

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?

Re: a (simple, but fun) riddle
 

I think he's right, although it probably doesn't matter how you make the initial division.

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! :D

Re: a (simple, but fun) riddle
 

Words fail me.

Re: a (simple, but fun) riddle
 

The title said it was a fun riddle...
I want my money back!

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?

Re: a (simple, but fun) riddle
 

Yes

Re: a (simple, but fun) riddle
 

No. You've removed two squares of the same colour, whereas a domino must cover a black one and a white one.

Re: a (simple, but fun) riddle
 

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

Re: a (simple, but fun) riddle
 

winner :)

Re: a (simple, but fun) riddle
 

God, I hate riddles.

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?

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.

Re: a (simple, but fun) riddle
 

You can't mix the bags.
You are allowed to take the beads out of the bags.

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.

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.

Re: a (simple, but fun) riddle
 

Don't take the risk, shoot yourself.

Re: a (simple, but fun) riddle
 

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.

Re: a (simple, but fun) riddle
 

n beads from the nth bag. The rest is really obvious.

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

Re: a (simple, but fun) riddle
 

You shoot one in the head, and just beat to death the other with the handle.

That is correct.

Re: a (simple, but fun) riddle
 

Neither am i :( basically i wanted a 'maths statement that encompassed the process' somewhere beyond a proof.

Re: a (simple, but fun) riddle
 

no it won't take off!!!