Riddle/Logic problem.

[royaldank]

This was posted at another site I read this morning. Thought it was interesting.
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You are sitting blindfolded at a table with some large number n of coins (n > 10) on it.
Each coin has a heads side and a tails side.
You are told that exactly 10 of the coins are currently heads up.
You are wearing gloves that will let you pick up, flip or move a coin, but not feel any differences in texture between sides.

Please make two groups of coins that contain exactly the same number of Heads-up coins.

You cannot see the coins, nor distiguish heads versus tails through touch.

 

[tweakmm]

Are there coins that are tails up?

 

[KoolAidKid]

Put all of the coins on their sides (rims) and make two groups.

 

[royaldank]

10 are heads up. The rest are tails up.

 

[Demon-Xanth]

Make three groups. One pile with all the coins, and two piles that are empty.

 

[amish]

stack them?

 

[dsfunk]

first you count them

 

[royaldank]

Originally posted by: dsfunk
first you count them

You really don't need to count them. You don't stack them, and there are no empty piles.

 

[chuckywang]

count them, split them in half to form two piles, and then flip all the coins in one pile.

 

[nineball9]

Pull out 10 coins to another group and flip them over.

 

[Zanix]

Originally posted by: royaldank

Originally posted by: dsfunk
first you count them

You really don't need to count them. You don't stack them, and there are no empty piles.

> than n piles though?

 

[element]

take 10 coins, set them aside and flip them all over. Leave the other pile as it is. There now you have the same number of heads in each pile.

 

[allisolm]

Originally posted by: chuckywang
count them, split them in half to form two piles, and then flip all the coins in one pile.

I don't think that works unless there are 20 coins.

 

[royaldank]

Originally posted by: nineball9
Pull out 10 coins to another group and flip them over.

Good job. You are correct.

 

[chuckywang]

Originally posted by: element
take 10 coins, set them aside and flip them all over. Leave the other pile as it is. There now you have the same number of heads in each pile.

damn, i was thinking along the same lines of thought...didn't get it though

 

[ArmenK]

Originally posted by: royaldank

Originally posted by: nineball9
Pull out 10 coins to another group and flip them over.

Good job. You are correct.

Doesnt work if you only have 10 coins since you said you cant have empty piles.

Edit: So you DO have to count them. If there are only 10 the solution is to split all the coins into two piles.

 

[Zanix]

Originally posted by: royaldank

Originally posted by: nineball9
Pull out 10 coins to another group and flip them over.

Good job. You are correct.

I don't understand.

 

[royaldank]

Originally posted by: ArmenK
Edit: So you DO have to count them. If there are only 10 the solution is to split all the coins into two piles.

My fault, I guess it needs to have n>10 in the problem. I copied and pasted from another site.

For those that might not understand, take n = 1000. Pull out 10 (1 is a head). Flip these 10 coins over (9 are now heads). At this point, you have 9 heads in the initial pot, and 9 heads in the group you flipped.

 

[nineball9]

Originally posted by: Zanix

Originally posted by: royaldank

Originally posted by: nineball9
Pull out 10 coins to another group and flip them over.

Good job. You are correct.

I don't understand.

Try it! (Or do it mentally).

Suppose you pull 2 Heads. That leaves 8 heads in the other pile. Now flip the 2-head pile and you now have 2 tails and 8 heads! (works for any permutation).

 

[Viper GTS]

Originally posted by: Zanix

Originally posted by: royaldank

Originally posted by: nineball9
Pull out 10 coins to another group and flip them over.

Good job. You are correct.

I don't understand.

Let's say there are 21 coins.

You pull 10, & happen to end up with 8 of the heads (leaving two in the larger pile of 11 coins).

You flip all 10, now you have 2 heads + 8 tails.

2 heads in each pile.

As long as you pull + flip the same number of coins as there are known heads it works.

Viper GTS

 

[Kev]

more problems please!

 

[shuan24]

Ohh the piles don't have to be equal...doh!

 

[SagaLore]

Originally posted by: royaldank
This was posted at another site I read this morning. Thought it was interesting.
------------
You are sitting blindfolded at a table with some large number n of coins (n > 10) on it.
Each coin has a heads side and a tails side.
You are told that exactly 10 of the coins are currently heads up.
You are wearing gloves that will let you pick up, flip or move a coin, but not feel any differences in texture between sides.

Please make two groups of coins that contain exactly the same number of Heads-up coins.

You cannot see the coins, nor distiguish heads versus tails through touch.

Well the riddle doesn't say that the coins needed to make the groups have to match heads or tails, just that you need the number of them. So you move all of the coins into the pile, and count to 10 moving a coin from one pile to the other, and then do it again for the 2nd group. Problem solved.

 

[SagaLore]

Originally posted by: royaldank

Originally posted by: ArmenK
Edit: So you DO have to count them. If there are only 10 the solution is to split all the coins into two piles.

My fault, I guess it needs to have n>10 in the problem. I copied and pasted from another site.

For those that might not understand, take n = 1000. Pull out 10 (1 is a head). Flip these 10 coins over (9 are now heads). At this point, you have 9 heads in the initial pot, and 9 heads in the group you flipped.

I don't see how that completes this:

Please make two groups of coins that contain exactly the same number of Heads-up coins.

That statement either means you just need 10 coins, or it means that you need at least 10 coins with the heads up. If the second is true, then having a group with 9 heads doesn't solve this.

 

[royaldank]

Originally posted by: SagaLore

Originally posted by: royaldank
This was posted at another site I read this morning. Thought it was interesting.
------------
You are sitting blindfolded at a table with some large number n of coins (n > 10) on it.
Each coin has a heads side and a tails side.
You are told that exactly 10 of the coins are currently heads up.
You are wearing gloves that will let you pick up, flip or move a coin, but not feel any differences in texture between sides.

Please make two groups of coins that contain exactly the same number of Heads-up coins.

You cannot see the coins, nor distiguish heads versus tails through touch.

Well the riddle doesn't say that the coins needed to make the groups have to match heads or tails, just that you need the number of them. So you move all of the coins into the pile, and count to 10 moving a coin from one pile to the other, and then do it again for the 2nd group. Problem solved.

?? (it simply means each group has to have the same number of coins facing heads up; 0 -> 10 of the coins can be face up)