Brain Exercise...

AoM · Postby **AoM** » Tue Nov 08, 2005 7:17 am

Someone besides BlueNine or Sho, riddle me this...

A man gives you a perfectly balanced scale and 10 gold coins. He tells you that one of the coins is a fake, whose weight is only slightly heavier than the other 9 coins. He says that you may only use the scale twice. If you can identify the fake golden coin with 100% certainty, he will let you keep the real gold coins, and the fake one too as a souvenir. Can it be done? How?

Sho · Postby **Sho** » Thu Nov 10, 2005 12:22 am

As they say in the IRC, you killed it!

Okay, so it's a bump.

Spider · Postby **Spider** » Thu Nov 10, 2005 12:38 am

Sorry AoM...I can only get it to 50 50...So, my answer is there is no answer for that. How about an answer and we move on?

Pie · Postby **Pie** » Thu Nov 10, 2005 2:21 am

IVE GOT IT!!!!! well, almost. I need three tries.

IT CANT BE DONE!!!

do i win?

El_Skwidd · Postby **El_Skwidd** » Thu Nov 10, 2005 3:34 am

Maybe you could add the coins one at a time, one to each side. Like this:

Coin A gets put on side 1.
Coin B gets put on side 2.

Neither A or B are fakes, so the scale is balanced..

Coin C gets put on side 1.
Coin D gets put on side 2.

The scale tips towards side 2. Coin D is the fake. Does that count as one use or two?

Spider · Postby **Spider** » Thu Nov 10, 2005 3:50 am

Thats two uses...And that would be luck if it happen, we're looking for 100% certainty

AoM · Postby **AoM** » Thu Nov 10, 2005 4:08 am

I messed up... I'm surprised some of the more knowledgable number theorists didn't correct me yet...

There are only 9 gold coins. One of them is a fake. If there were 10 gold coins, you would indeed require three uses of the balance. Now you can answer the question.

*hides from various large objects thrown his way*

Sho · Postby **Sho** » Thu Nov 10, 2005 4:19 am

I figured this was just a modification of the problem, requiring you to prove that 10 is not doable. Besides, I'm a nonparticipant observer.

AoM · Postby **AoM** » Thu Nov 10, 2005 4:23 am

I asked "Is this possible?" as my first question, so my riddle has been legit. But putting up an unsolvable riddle and expecting them to disprove it would be something kind of... well, ass-holish. So I will be honest... I had meant to put up 9 instead of 10.

But it is still up to the others to explain how and why.

Spider · Postby **Spider** » Thu Nov 10, 2005 4:25 am

*sighs as he answers* Alright, divide the coins into 3 sets of 3. Place one set on one side and any other on the other side. If they are equal, then the coin is in the other set of 3 coins, take 2 from those 3 and balance them, if equal, your coin is the one that hasn't been measured.

If any time one is heavyier, like the first measurement, then you have your other set to measure. Hope you understand that...And someone else make a riddle, I don't like the ones I can think of....

Pie · Postby **Pie** » Fri Nov 11, 2005 12:31 am

WAIT!!!! THATS NOT FAIR!!!! I SAID

IT CANT BE DONE!!!

do i win?

FIRST!!! I SHOULD WIN!!!!

My answer was to put 5coints on one side and 5 on the other. and then i would put three one one side, and three (one being one that i know is real) on the other side and balanc it.

that gets us to a chance of two, or three. So, i will take one coin frome one side, and mesure the other two.

Basicly the same, exept i started with five.

AoM · Postby **AoM** » Fri Nov 11, 2005 3:50 am

Pie, your reasoning was flawed, so no, you don't win. But since Spider isn't posting another riddle, I suppose anyone is allowed to suggest one.

Sunni Daez · Postby **Sunni Daez** » Fri Nov 11, 2005 12:00 pm

Well, I didn't answer a riddle, but I have one....

Three men, returning from a hunting trip, stopped at a motel for the night. The clerk charged them 30$ for the room, so each man gave 10$.
The Clerk realized he had over charged the men by 5$ So he went to thier room to give a refund. Well, not knowing how to divide the money equally, he gave each of the three men 1$ and kept the other 2$...

Now, this means each of the three men only paid 9$...

3x9$ = 27$, the Clerk kept 2$..... 27$ +2$ = 29$ where did the other Dollar go?

SekoETC · Postby **SekoETC** » Fri Nov 11, 2005 12:54 pm

Well... it's sure twisting my brain... but if he overcharged them $5 that means the real price of the room was $25, and since the men paid $27, there's the $2 over that goes to the clerk to make it 25. Nothing vanished, it's just not supposed to be count that way. There's $25 in the cash register, 3*$1 in the men's pockets and $2i in the clerk's pockets, totalling in $30.

Postby **Jur Schagen** » Fri Nov 11, 2005 1:17 pm

There are four people on the side of a river, and one boat. The boat can only carry two people, and all want to cross. Agar can cross in 1 minute, Bran in two, Chris in five and Doug in 10 minutes. For some reason, the boat speed is limited to the slowest of the people in it. (if Doug sails with Agar, crossing takes ten minutes); and of course, someone has to sail it back too. What is the shortest time to get the group across?

It is said this one is used in the selection procedure for a job at Microsoft...

Edit - no, the answer isn't 19.

Brain Exercise...

Who is online