coins in a dark room
 

Re: coins in a dark room
 

I think you grab 25 coins and flip over 7 of them

have to work this out

Re: coins in a dark room
 

actually, i think it might be the other way around; grab 25, and flip 18.

i outta be studying for a fluid mech. exam, but this riddle keeps distracting me...

Re: coins in a dark room
 

{Ken starts warming up his vintage Cray 1}

Re: coins in a dark room
 

Choose a common coin type, let's say you feel US dimes. Put the dimes in one group and the rest of the coins in a different group. Bingo, 2 groups of coins :p
EDIT: Oops, forgot about the heads up part.

Re: coins in a dark room
 

Stand them on edge.

Re: coins in a dark room
 

Jack has left the box :^)

some coins wont stand on their edge

how bout, put 25 in your left pocket, and 25 in your right pocket?

(no heads are showing in either pocket)

Re: coins in a dark room
 

Ken took the money and ran^

The room is dark -> group anyway you want - neither group has any heads showing : zero == zero

Re: coins in a dark room
 

as you pick up each coin, cut it in half

put the two halfs in separate piles

or better yet, you get an elephant, a bowling ball, and 4 pounds of wax...

Re: coins in a dark room
 

lol
i don't know the answer either
its bothering me

Re: coins in a dark room
 

this is my standard answer for all brainstorming sessions, no matter what the question or problem is

Re: coins in a dark room
 

Why don't you just turn on the light, and then flip over the coins?

Re: coins in a dark room
 

because it defeats the object of the excercise
lol

Re: coins in a dark room
 

I knew that, but then Ken said something about some elephant so I thought I was entitled to say something about the lights. :)

Was this a homework assignment or something?

Re: coins in a dark room
 

I think I reasoned it out, I had the right idea on my first post, but the wrong number

the original pile has 32 heads up, so you take 32 from the pile and flip them over. That is your second pile.

If you grabbed all 32 heads then there are no heads left in either pile

if you grabbed 18 heads and 18 tales, then you left 18 heads in the first pile, and when you flip the second pile you have 18 heads in both piles

if you grabbed 19 heads and 17 tales, then you left 17 heads in the first pile, and when you flip the second you have 17 heads and 19 tails

it works for all combinations

but I like the pocket answer better :^)

Re: coins in a dark room
 

Ken,

Last time I checked, 18+18=36 not 32.

Mike

Re: coins in a dark room
 

Your answer is right but the explanation is wrong:

The original piles has 32 heads up, so you take 32 from the pile and flip them over. That is your second pile.

If you grabbed all 32 heads then there are no heads left in either pile

if you grabbed 18 heads and 14 tales, then you left 14 heads in the first pile, and when you flip the second pile you have 14 heads in both piles

if you grabbed 19 heads and 13 tales, then you left 13 heads in the first pile, and when you flip the second you have 13 heads and 19 tails

it works for all combinations

Regards,

Mike

Re: coins in a dark room
 

Are you sure thats the riddle? I did a bunch of googling (Try googling a riddle :rolleyes: ) so yeah here is what i came up with.
"You are wearing a blindfold and thick gloves. An infinite number of quarters are laid out before you on a table of infinite area. Someone tells you that 20 of these quarters are tails and the rest are heads. He says that if you can split the quarters into 2 piles where the number of tails quarters is the same in both piles, then you win all of the quarters. You are allowed to move the quarters and to flip them over, but you can never tell what state a quarter is currently in (the blindfold prevents you from seeing, and the gloves prevent you from feeling which side is heads or tails). How do you partition the quarters so that you can win them all?"
uh i also found a forum based on this riddle. http://www.ocf.berkeley.edu/~wwu/cgi...m=102 8100705

Re: coins in a dark room
 

it was actually just something that my friend said to me
i don't have a clue where he got it from

Re: coins in a dark room
 

Which one of the states is infinately large? :-P
What state is on the back of all the quarters?
Do you know what breed of horse is on the back of the Delaware quarter?



...A quarterhorse.

Re: coins in a dark room
 

Here's the algebraic proof that the solution of grabbing 32 coins and flipping them works.

Let's call our pile of 18 Pile I
Let's call our pile of 32 Pile II

Let a be the number of heads in Pile I
Let b be the number of tails in Pile I
Let x be the number of heads in Pile II
Let y be the number of tails in Pile II

Therefore,

eq1: a + b = 18 (From Pile I)
eq2: x + y = 32 (From Pile II)
eq3: a + x = 32 (From the total number of heads)
eq4: b + y = 18 (From the total number of tails)

Combining eq2 & eq3 we see that,
y - a = 0 --> y = a

Therefore the number of heads in Pile I is the same as the number of tails in Pile II

Now, we flip all 32 coins in Pile II. So all the heads become tails and vice-versa. So now we have y heads and x tails in pile II.

Since a = y, we see that the number of heads in both piles I & II is equal!

Re: coins in a dark room
 

ok, I was in a hurry: 32 / 2 = 16, not 18

32 heads and no tails
31 heads and 1 tail...
down to 18 tails and 14 heads, it comes out the same - no matter which 32 you grab

for the question with infinity coins, you grab 20 and flip them, and make that your second pile

I want to know, how can you tell only 20 were originally tails up, if there is an infinite number of coins in the room?

Re: coins in a dark room
 

You can't, as you can't fit an unlimited amount of something in a finite space.

Not to mention that there is no such thing as infinity coins.