interesting math riddle

[furyshade]

my friend showed me this today, he doesn't know the answer and it has been bugging me, anyone wanna give a shot at it? please dont just look up answer, and if you know it don't post it. oh and my friend said it is a math riddle, not some trick like "turn on the light".

There are 50 coins, 18 heads up, 32 tails up in a dark room. You want to make two piles that have an equal number of heads. (You don't care what size these piles are, and the piles don't necessarily have to be the same size as each other. So, for instance, if you had a way to verify that two coils were tails, you could put them in two separate piles of 1 coin each, and you'd be done.) You cannot feel the coins or anything like that. The only thing you can do is move them around and flip them.

[thylacine]

[ QUOTE ]
my friend showed me this today, he doesn't know the answer and it has been bugging me, anyone wanna give a shot at it? please dont just look up answer, and if you know it don't post it. oh and my friend said it is a math riddle, not some trick like "turn on the light".

There are 50 coins, 18 heads up, 32 tails up in a dark room. You want to make two piles that have an equal number of heads. (You don't care what size these piles are, and the piles don't necessarily have to be the same size as each other. So, for instance, if you had a way to verify that two coils were tails, you could put them in two separate piles of 1 coin each, and you'd be done.) You cannot feel the coins or anything like that. The only thing you can do is move them around and flip them.

[/ QUOTE ]

I got it. Posting the answer in white below.

<font color="white"> Make a pile of 18 and a pile of 32. Then turn the pile of 32 upside-down.
. </font>

[bigpooch]

I hadn't seen this before, but it's quite easy once you see
the "light" (answer in white):

<font color="white">
Take 18 of the 50 coins and flip each of these eighteen to
the other side. The flipped pile now has the same number of
heads as the unflipped pile.

If there were x heads in an unflipped pile of y coins and
you flip the other 50-y coins to the opposite side, the
number of tails in the pile of 50-y coins before flipping
is 50-y-(18-x) since there are 18-x heads in the pile of
50-y coins. For x and 50-y-(18-x) to be equal, you simply
need 32-y = 0 or y = 32 or the number of unflipped coins to
be 32.

Obviously, this generalizes to a room full of some number of
coins (you don't even need to know how many) where you know
the number of coins that are heads. If you know there are
N heads in the room, just take N of the coins and flip each
of them. The flipped pile will then have the same number of
heads as the unflipped pile.

</font>

[furyshade]

[ QUOTE ]
[ QUOTE ]
my friend showed me this today, he doesn't know the answer and it has been bugging me, anyone wanna give a shot at it? please dont just look up answer, and if you know it don't post it. oh and my friend said it is a math riddle, not some trick like "turn on the light".

There are 50 coins, 18 heads up, 32 tails up in a dark room. You want to make two piles that have an equal number of heads. (You don't care what size these piles are, and the piles don't necessarily have to be the same size as each other. So, for instance, if you had a way to verify that two coils were tails, you could put them in two separate piles of 1 coin each, and you'd be done.) You cannot feel the coins or anything like that. The only thing you can do is move them around and flip them.

[/ QUOTE ]

I got it. Posting the answer in white below.

<font color="white"> Make a pile of 18 and a pile of 32. Then turn the pile of 32 upside-down.
.

[/ QUOTE ]

this doesn't make sense, if you have a pile of 32 that is half heads and half tails, the flipped pile will be the same, the coins aren't already in piles, that would be stupidly easy
</font>

[thylacine]

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I got it. Posting the answer in white below.

<font color="white"> Make a pile of 18 and a pile of 32. Then turn the pile of 32 upside-down.
. </font>

[/ QUOTE ]

<font color="white"> this doesn't make sense, if you have a pile of 32 that is half heads and half tails, the flipped pile will be the same, the coins aren't already in piles, that would be stupidly easy
</font>

[/ QUOTE ]

Huh? Explain.

[billygrippo]

heh i saw this problem last night. it was worded funny tho and i thought both piles had to be equal size.

[furyshade]

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I got it. Posting the answer in white below.

<font color="white"> Make a pile of 18 and a pile of 32. Then turn the pile of 32 upside-down.
. </font>

[/ QUOTE ]

<font color="white"> this doesn't make sense, if you have a pile of 32 that is half heads and half tails, the flipped pile will be the same, the coins aren't already in piles, that would be stupidly easy


[/ QUOTE ]

Huh? Explain.

[/ QUOTE ]
why would making two piles and flipping one over do anything, the piles aren't sorted so the piles could and probably do contain boths heads and tails coins, so doing this accomplishes nothing
</font>

[Philo]

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I got it. Posting the answer in white below.

<font color="white"> Make a pile of 18 and a pile of 32. Then turn the pile of 32 upside-down.
. </font>

[/ QUOTE ]

<font color="white"> this doesn't make sense, if you have a pile of 32 that is half heads and half tails, the flipped pile will be the same, the coins aren't already in piles, that would be stupidly easy
</font>

[/ QUOTE ]

Huh? Explain.

[/ QUOTE ]

<font color="white">He's saying that this won't work b/c if the pile of 32 happens to have 16 heads and 16 tails, then after flipping them that pile will still have 16 heads and 16 tails, so nothing has changed, so that can't work. </font>

[billygrippo]

ill explain in white.

<font color="white"> we know there are 18 heads. we take 18 coins randomly and put them into a pile. we flip the coins. now lets say that we took 10 heads, leaving 8 in the other pile. once they are flipped the 10 heads become tails and the 8 tails become heads.

lets do it again. this time we get 1 heads in our 18 pile, leaving 17 heads in the other pile. flip the 18 and now we have 17 heads.

as long as u take the same amount of coins as tehre are heads and flip them it will always work. </font>

[furyshade]

[ QUOTE ]
ill explain in white.

<font color="white"> we know there are 18 heads. we take 18 coins randomly and put them into a pile. we flip the coins. now lets say that we took 10 heads, leaving 8 in the other pile. once they are flipped the 10 heads become tails and the 8 tails become heads.

lets do it again. this time we get 1 heads in our 18 pile, leaving 17 heads in the other pile. flip the 18 and now we have 17 heads.

as long as u take the same amount of coins as tehre are heads and flip them it will always work. </font>

[/ QUOTE ]

something about this im missing, are the two steps you give separate or consecutive?

[bunny]

<font color="white">Make a pile of 18 random coins - there are x heads in it.
This means pile one has x heads and 18-x tails.
Pile two has 18-x heads and 32 -(18-x) tails.
If you now flip pile one it has x tails and 18-x heads.
Both piles now have the same number of heads. </font>

[jay_shark]

[ QUOTE ]
my friend showed me this today, he doesn't know the answer and it has been bugging me, anyone wanna give a shot at it? please dont just look up answer, and if you know it don't post it. oh and my friend said it is a math riddle, not some trick like "turn on the light".

There are 50 coins, 18 heads up, 32 tails up in a dark room. You want to make two piles that have an equal number of heads. (You don't care what size these piles are, and the piles don't necessarily have to be the same size as each other. So, for instance, if you had a way to verify that two coils were tails, you could put them in two separate piles of 1 coin each, and you'd be done.) You cannot feel the coins or anything like that. The only thing you can do is move them around and flip them.

[/ QUOTE ]

When I first read this problem , I immediately thought of a related old problem : Show that if you cut a deck of cards in half , then the number of red cards in one pile is equal to the number of black cards in the other . Using this insight , we can grasp a solution .

Instead of dividing the coins in 2 equal piles , we divide the coins so that one pile has 18 random coins and the other has 32 .

Let a be the number of heads in pile 1(18 cards)
Let b be the number of tails in pile 1 .

a+b=18
18-a + 32-b =32

So the number of tails in pile 1 is equal to the number of heads in pile 2 . Now all you need to do is flip all the coins from pile 1 .

[thylacine]

OP, let us know when the penny drops.

[GoRedBirds]

I saw this same riddle (different numbers) somewhere a couple days ago. Someone else said they saw it too. Any idea where we saw it? I'm drawing a blank.

[madnak]

I think you confused him, thylacine. Your answer is wrong:

<font color="white">You have to flip the 18, not the 32.

If you flip the 32 you'll always end up with more heads in that pile. The 18 pile will have x heads and 18-x tails, the 32 pile has 18-x heads and 32-(18-x) tails. Flip the 32 and you now have 18-x tails and 32-(18-x) heads. So x heads in the 18 pile and 32-(18-x) heads in the 32 pile. That's x and 14+x, the 32 pile will always have 14 more heads than the 18 pile.</font>

[thylacine]

[ QUOTE ]
I think you confused him, thylacine. Your answer is wrong:

<font color="white">You have to flip the 18, not the 32.

If you flip the 32 you'll always end up with more heads in that pile. The 18 pile will have x heads and 18-x tails, the 32 pile has 18-x heads and 32-(18-x) tails. Flip the 32 and you now have 18-x tails and 32-(18-x) heads. So x heads in the 18 pile and 32-(18-x) heads in the 32 pile. That's x and 14+x, the 32 pile will always have 14 more heads than the 18 pile.</font>

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Ok, I typed quickly thinking it started with 32 heads, 18 tails. Anyway, there are about ten explanations by now.