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Brainteaser, anyone?

     
12:49 pm on Dec 24, 2002 (gmt 0)

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Hello, im bored its christmas eve im at work, and im work again on boxing day, so i thought id post this, although no-one will answer it because yopu have better things to do... like... enjoy christmas.

Two guys are playing chess they both play ten games, and win 10 each, they do not draw, tie or whatever. how is this possible?

Any thoughts.....

5:09 pm on Dec 26, 2002 (gmt 0)

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White,

This geometry only works if his cabin is at one of the poles.

Jim

5:12 pm on Dec 26, 2002 (gmt 0)

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Do polar bears live at the South Pole?

lawman

5:13 pm on Dec 26, 2002 (gmt 0)

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a white bear of course.
5:29 pm on Dec 26, 2002 (gmt 0)

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> Do polar bears live at the South Pole?
No
5:38 pm on Dec 26, 2002 (gmt 0)

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OH! There is one of those riddles I remember from HS. I could never got it.

You have a lion, lamb, wolf, and a head of cabbage (or something like that)that you have to get accross the river. They all have to be taken accross without getting eaten by one another and you have to have one in the boat at all times.

5:46 pm on Dec 26, 2002 (gmt 0)

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first take the lamb.
Then bring back the lamb and take the cabbage.
Then bring back the cabbage and take the wolf.
Then bring back the wolf and take the Lion.

One at a time. Always ONE on the boat.

6:47 pm on Dec 26, 2002 (gmt 0)

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I dont think that is it.

"first take the lamb.
Then bring back the lamb and take the cabbage.
Then bring back the cabbage and take the wolf.
Then bring back the wolf and take the Lion.
One at a time. Always ONE on the boat."

step one ok.
step two while lamb is on shore the Lion and wolf eat the lamb.

I could be wrong this one has been chewing on me for about 12 years.

8:24 pm on Dec 26, 2002 (gmt 0)

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this one has been chewing on me for about 12 years.

It's probably been haunting on you for 12 years because it can't be solved in the way you know it. ;)

It's a classic logic problem, with a few of the details jumbled...

It's known under a few names, the most popular being "The Wolf, the Goat, and the Cabbage", or "The Fox, The Goose, and The Bag of Corn".

I don't believe there's anywhere in the world that lions and wolves share a habitat. :)

The farmer needs to get all three across a river and can only fit one in the boat at a time. The goat (goose) will eat the cabbage (corn) and the wolf (fox) will eat the goat (goose) if they are left unattended.

Now, with a bit of ingenuity you should be able to solve the problem.

9:15 pm on Dec 26, 2002 (gmt 0)

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first take the goat. Leave it on shore.
Then, take the fox to shore but bring back the goat.
Finally, leave the goat on the original shore and take the cabbage to where the fox is.

Last trip. Bring back the goat where the fox and cabbage is.

I believe there is no Lion in the original game.

FOX
GOAT
CABBAGE

.....

FOX GOAT
CABBAGE
.....
CABBAGE FOX
GOAT (PICKED UP)

.....
GOAT (LEFT) FOX
CABBAGE

.....
FOX
CABBAGE
GOAT

There ya go.

oops, it didn't take the table to view the placements...

10:10 pm on Dec 26, 2002 (gmt 0)

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Yeah the old memory does not serve me like it should, but now it makes sence. I threw in the Lion and the lamb due to reading the Old Testament analogies. Some times the wires get crossed.

Oh I actually know the next one.

Using a 3 gallon and a 5 gallon water jug. You need to get 2 gallons in the 5 gallon jug.

This one is fairly easy.

[edited by: miles at 10:13 pm (utc) on Dec. 26, 2002]

10:13 pm on Dec 26, 2002 (gmt 0)

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Fill the 5 gal. jug, then fill the 2 gal. jug out of the 5 gal. one, and you'll have 3 gal. left in it.
11:00 pm on Dec 26, 2002 (gmt 0)

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John's mother has 3 children, Penny, Nicholas, and?
11:38 pm on Dec 26, 2002 (gmt 0)

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John?
1:48 am on Dec 27, 2002 (gmt 0)

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OK, this one's simple:

How come a man can't marry his widow's sister?

1:49 am on Dec 27, 2002 (gmt 0)

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Because a dead man can't marry anyone. :) (I'm doin' great on the easy ones!)
1:52 am on Dec 27, 2002 (gmt 0)

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OK, not a riddle but just a bad joke (right up mivox's alley):

Did you hear about the Eskimo that was stabbed to death with the icicle?

*taps fingers waiting for a response...*

1:55 am on Dec 27, 2002 (gmt 0)

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I heard a riddle like that once... it goes through a whole elaborate murder/suicide scenario, and the solution to the riddle (what happened to the murder weapon, if everyone involved was dead?) was that the victim was stabbed with an icicle, and it melted.

But I can't think of a punchline for it. :(

1:57 am on Dec 27, 2002 (gmt 0)

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Ha! got you!

>Did you hear about the eskimo that was stabbed to death with the icicle?

He died from COLD CUTS!

:)

1:58 am on Dec 27, 2002 (gmt 0)

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*groan* whooo.... I'm relieved I couldn't think of the punchline now. ;)
2:16 am on Dec 27, 2002 (gmt 0)

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Extensive scientific research has demontrated that there is a direct relationship between foot size and performance in children's spelling bees. What is the explanation for this?
2:30 am on Dec 27, 2002 (gmt 0)

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Bigger the foot, the older the child. The older the child, the better the performance.

Now a real brain teaser. It's been several years since I've done this one, so I'll have to sit down and figure it out again:

You have 12 coins that appear identical, and a set of scales. One of the coins is either lighter or heavier than the other 11. Using only three weighings, how can you tell which coin is different and whether it is heavier or lighter.

2:48 am on Dec 27, 2002 (gmt 0)

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Number the coins 1 through 12.

Perform the following three weighings:

Weighing 1 - Left side: 10 3 2 1 Right Side: 11 6 5 4
Weighing 2 - Left side: 11 3 2 1 Right Side: 7 8 10 9
Weighing 3 - Left side: 1 4 10 7 Right Side: 6 3 9 12

If the left side is most heavy call it "L".
If the right side is most heavy call it "R".
If the left and right sides have the same weight call it "B".

These are the possible outcomes:

LLL 1 heavier
LLR 3 heavier
LLB 2 heavier
LRL 10 heavier
LRB 11 lighter
LBL 6 lighter
LBR 4 lighter
LBB 5 lighter
RLR 10 lighter
RLB 11 heavier
RRL 3 lighter
RRR 1 lighter
RRB 2 lighter
RBL 4 heavier
RBR 6 heavier
RBB 5 heavier
BLL 9 lighter
BLR 7 lighter
BLB 8 lighter
BRL 7 heavier
BRR 9 heavier
BRB 8 heavier
BBL 12 lighter
BBR 12 heavier

[edited by: Dante_Maure at 2:59 am (utc) on Dec. 27, 2002]

2:55 am on Dec 27, 2002 (gmt 0)

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Using only a 6 minute hourglass and an 13 minute hourglass how can you accurately measure 20 minutes?
3:15 am on Dec 27, 2002 (gmt 0)

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I'll take a shot...

Begin with both hourglasses empty on top.
Flip both hourglasses.
Allow the 6-minute hourglass to run out, then flip it over immediately.
When the 6-minute hourglass runs out again, leave it, but begin timing 20-minute interval.
When the 13-minute hourglass runs out (one minute into the 20-minute interval), flip it over.
When the 13-minute hourglass runs out a second time (14 minutes into the 20-minute interval), flip the 6-minute hourglass over.
It will run out at the 20-minute mark.

Jim

3:17 am on Dec 27, 2002 (gmt 0)

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Wow Dante, I just solved the coin scenario and I have no idea what your solution is. Let me ruminate awhile.

lawman

3:44 am on Dec 27, 2002 (gmt 0)

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Dante:

I must apologize. After looking at my solution, it appears just as confusing as yours. Here it is FWIW:

Divide coins into three stacks of 4.

Put 4 coins on one side of scales and 4 coins on other.

First Possibility:

The two weighed stacks are equal and then the odd coin is in the stack of 4 that was not weighed.

Take 3 coins from the eight you've already weighed and put on one side. Then three coins from unweighed

stack and put on the other. If they are equal, then you have narrowed it to one coin. Simply place it on one side

of the scale along with one of the known coins to determine if it is lighter or heavier.

However, if they are unequal, then you will know if it is lighter or heavier. Simply place one of the three coins on

one side and one on the other. If one is lighter, it will rise. If they are equal, then it is the remaining coin.

Second Possibility:

The two weighed stacks are unequal and you've narrowed it down to eight. You know either that the lower stack

has a heavier coin or the higher stack has a lighter coin.

Place 3 coins from the lower stack and 1 coin from the higher stack on the first side. Place 3 known coins and

the remaining 1 coin from the lower stack on the second side.

If the first side goes down, you have narrowed it to on of the 3 coins from the original lower stack. Simply place

one of each on either side of the scale. If one side goes down, that is your coin. If neither goes down, it is the

remaining coin.

If the second side goes down, you have narrowed the search to the 1 coin from the original higher stack on the

first side or the 1 coin from the original lower stack on the second side. Simply weigh one of them agains one

of the know coins for the answer.

If both sides are equal, then you know it is one of the three remaining coins from the higher (lighter) stack.

Simply place one on one side and one on the other. If they are equal, then it is the remaining coin.

I hope this is easier to follow than Dante's.

lawman

6:16 am on Dec 27, 2002 (gmt 0)

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I must apologize. After looking at my solution, it appears just as confusing as yours.

Heh. With this problem the only folks likely to be anything other than confused carry Phd's in Mathematics. I posted a solution, and I still find it confusing. ;)

The last post on this page [mathforum.org] goes into a much more thorough explanation of the systematic approach I used above.

6:19 am on Dec 27, 2002 (gmt 0)

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Way to go jdMorgan!

Now, the question is whether you or any other members can show a solution which can be executed in less time. (yours would take 32 minutes altogether)

6:30 am on Dec 27, 2002 (gmt 0)

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Start both hourglasses simultaneously. When the 6 minute hourglass has ended, flip the 13 minute hour glass and let it run for the remaining 7 minutes. Flip it once again for 13 more minutes to equal the 20 minutes. 26 minutes total. Can it be done quicker?
7:23 am on Dec 27, 2002 (gmt 0)

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Bingo coconutz. :)

That's the fastest solution I've been able to work out myself.

This 196 message thread spans 7 pages: 196