Riddle

Turner · Oct 12, 2005

Hit the quote button under the post you wish to quote, or put in [quote=username][/quote].

Actually, with the 12 coins you'd have to do it on each side of the balance. Since 23 are the same weight, and one isn't, the side with the lighter coin would raise up. Remove the 12 coins that you know now are of equal weight, and split the 12 coins and repeat until you have the lightest coin.

Trogg · Oct 12, 2005

Turner_727 said:
Actually, with the 12 coins you'd have to do it on each side of the balance. Since 23 are the same weight, and one isn't, the side with the lighter coin would raise up. Remove the 12 coins that you know now are of equal weight, and split the 12 coins and repeat until you have the lightest coin.

Thanks for telling me, i actually thought i had to type username not the actual name :lol:

Sorry no. what you do is you weigh 8 coins against 8 coins, then from the pile of 8 that has the gold one in it you weigh 6, and after weighing 6 either one of the groups of 3 weighed or the group of 2 woudl have the godl coin, if its in the group of three weigh one of them against the other and one will be the godl coin or else the one you didn't weigh is :crazyeye:

CivFan91 · Oct 12, 2005

But Turner's would work! And it's easier to read.

Perfection · Oct 12, 2005

Now how do you do 13?

seasmath · Oct 12, 2005

Weigh 6 and 6. If equal weight, the one left out is the answer. If uneven, split the lighter one into 3 and 3. Split lighter one into 1 and 1. If equal, the third one is the answer. If uneven, the lighter one is the answer...

Turner · Oct 12, 2005

C3CFanatic0014 said:
But Turner's would work! And it's easier to read.

Suck up!

I left my riddles at home...gonna have to post some tomorrow.

CivFan91 · Oct 12, 2005

And proud of it!

Perfection · Oct 12, 2005

Here's a great coin question, you have a batch of coins, inside this batch a counterfiet one is either ligher or heavier. You have three weighings on a balance, what's the maximum number of coins you can have in the batch and still find the counterfeit? What if you have unlimited coins with the right weight as well?

CivFan91 · Oct 12, 2005

If you have unlimited coins with the correct weight, it's impossible.

WildWeazel · Oct 12, 2005

27 is the most.

Trogg · Oct 12, 2005

C3CFanatic0014 said:
But Turner's would work! And it's easier to read

actually it wouldn't because with his you weigh 6 against 6 that leaves 12 you didn't weigh to the side and you cna't find out which coin is which with only 2 tries on the balance for 12 coins :crazyeye:

i was a little complex on my explanation hope your brain didn't [pissed]

CivFan91 · Oct 12, 2005

Yeah, it did. :lol:

I love that smiley.

[pissed]

Perfection · Oct 13, 2005

C3CFanatic0014 said:
If you have unlimited coins with the correct weight, it's impossible.

They're not in the batch

Weasel Op said:
27 is the most.

how would you do it?

WildWeazel · Oct 13, 2005

weigh 2 sets of 9, take the [lighter/heavier] set or the other set if they're equal
weigh 2 sets of 3, same as above
weigh 2, if they're equal it's the other one

Trogg · Oct 15, 2005

NO new riddles?!?!?!? sigh the flare of this thread is gone now. I think there was one i did that no ones answered yet though.

WildWeazel · Oct 15, 2005

I have an answer for a riddle, but I don't know how to ask it. :crazyeye:

CivFan91 · Oct 15, 2005

Here's how to ask: 1) Type it up, 2) Hit 'Submit', and BINGO!

Turner · Oct 15, 2005

Sam and his friends were playing cards. Sam decided to see how bright they were. "In an ordinary deck of playing cards, two of the jacks have two eyes, and two of them have just one. How many eyes are on the four Jack cards?"

seasmath · Oct 15, 2005

Six? (xcl)

WildWeazel · Oct 15, 2005

12????????????

Riddle

Deity

Spiritual Entity

Emperor

The Great Head.

Bolos!

Deity

Emperor

The Great Head.

Emperor

Carthago Creanda Est

Spiritual Entity

Emperor

The Great Head.

Carthago Creanda Est

Spiritual Entity

Carthago Creanda Est

Emperor

Deity

Bolos!

Carthago Creanda Est

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