Thread: Post A Logic Puzzle

Correct! I have to find something harder. I knew that one would get ripped to sheards. LOL..You Guys.

Originally Posted by becomingagodo

Finally some assumption can be thrown away. Their two more how many bags can you weight if it all nine then it easy and can you take the coins out of the bag and what do you mean arrangment. If the coin were infintly small then it wouldnt be that hard to pick up. Also does the scale measure exaculy what condition is it in strength of gravity does the scale suffer from some sort of uncertainty princple i.e. is it moving and below plank constant.
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You are trying to make this way too complex 0.o

You can take any amount of coins from any or all of the nine bags and weigh them. Lets pretend this scale is a heavenly scale that takes perfect mesurements mmkay?

does it matter what the coins weigh? like say the gold ones weigh 1g and the lead weigh .5g. then you measure two and hope the total is 1.5... my head hurts.

Originally Posted by jacobo

does it matter what the coins weigh? like say the gold ones weigh 1g and the lead weigh .5g. then you measure two and hope the total is 1.5... my head hurts.
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It doesn't matter exactly how much the coins weigh. As long as the gold weight differnt from the iron. Giving the weights was just for mister infinate over there =P

Finally some assumption can be thrown away. Their two more how many bags can you weight if it all nine then it easy and can you take the coins out of the bag and what do you mean arrangment. If the coin were infintly small then it wouldnt be that hard to pick up. Also does the scale measure exaculy what condition is it in strength of gravity does the scale suffer from some sort of uncertainty princple i.e. is it moving and below plank constant.[/b]

Haha. You're not a high-brow thinker, you're a freak!

As for finding the iron coins I can't think of an answer. You can get the weight of both coins from weighing one gold coin, since the iron weighs half. But you only get one chance to weigh them, meaning that you'll have both weights but you won't actually be able to test anymore coins against those weights, i.e. impossible? You could try and weigh them with your hand but A) that would probably count as weighing too so you'd still violate the 1-weigh rule and B) the weights might be too close to each other to be easily discerned).

And besides which, you don't even know which coins are iron and which aren't, so the chances of you picking the right type of coin are about 88%, and you would waste your one go at weighing and screw it up the other 12% of the time. So even if you could get an answer this way, it wouldn't always work.

So:

• Weighing combinations of coins is going to be useless as you don't know which coin is which and can't separate the values from each other. You get one weigh, and you'd receive a number like 43 from, say, 8 gold coins and the 1 iron coin - how much does a golden coin weigh? (A formula: 8g + i = 43 - totally and utterly plum useless). If you weigh one coin (which you better hope is a gold one in the first place, I don't like the element of chance in this) you can find the weight of both but then you've used your one weigh.
• You can only weigh once and so you can't check the other bags anyway.
• You have no way of knowing what kind of coin you are weighing without weighing more than one type of coin, which - see above - is useless.

I must be missing something big here (and I blame Mr. Infinity, also known as becomingagodo over there for throwing uncertainity principles around like common gardening tools are often thrown in such topics) because I really can't see a clue on how you'd proceed. Unless this is a trick riddle and arby is going to get shot.

does it matter what the coins weigh? like say the gold ones weigh 1g and the lead weigh .5g. then you measure two and hope the total is 1.5... my head hurts.[/b]

Even if we knew the weights, we'd still be screwed by the element of chance. Whoever this wizard is, I know he wouldn't be my friend.

FURTHER THOUGHT: Can he throw them at the magician's head and see which coins cause the magician to collapse faster?

if every coin was infinitly small, with an infinit number of coins, it would be bloody heavy

i'm almost certain you need the original weight of one gold coin to do this...

i was thinking you could make (and label) stacks of coins from the different bags, each stack has a different number of coins, and keep track of what the weight 'should be' if all the coins were gold. then somehow through the magic of math narrow down the possible stacks according to what the weight actually came out to be... but i'm pretty sure that you would need to know the exact weight of a gold coin to do that... and you would also run into the problem if you choose an even number of coins for the lead stack... because then your weight wouldn't come out as being a decimal.... i don't know... my head hurts.

yeah that would work, if you knew the weight of one coin you could choose the arrangement 0,1,2,3,4,5,6,7,8.

If the first bag is iron, then everything you weigh is gold so the total weight is 36 times the weight of one gold coin.

If its in one of the others the total weight will be the that answer minus (N*W/2) where N is the number of coins you put in from that bag and W is the weight of the gold coin.

So to work out which bag you can just calculate the "missing weight" and divide by the weight of a coin to get the number of iron coins, and if each bag has a unique number of coins in the arrangement then you know where they came from. (This probably makes more sense in my head)

Yep, thats right =D

Now let me find an even harder one

Wait, so you mean the reason I didn't get the riddle at all was because:

The iron coins weigh half as much as the gold coins and the man may only weigh any arrangement of coin from any bags, once. (one total)[/b]

I took that to mean you can weigh the coins - any number you like - once. Are you meaning something else, or am I not understanding the solution as well as the problem?

The solution is, you take 1 coin from the first bag, 2 coins from the second, 3 from the third and so on. Then you put all the coins on the scale and weigh them. The weight will be off by a certain amount (lets say the gold each weighed 1g) all the coins would weigh 45g if they were all gold but they won't since one bag is iron. Lets say the weight ended up being 44g. Because we know that the iron weighs .5 less then the gold we can surmise that there are 2 iron coins on the scale. Therefore bag 2 is the culprit =D.

Originally Posted by jacobo

Three men in a cafe order a meal the total cost of which is $15. They each contribute $5. The waiter takes the money to the chef who recognizes the three as friends and asks the waiter to return $5 to the men.

The waiter is short on cash so instead of going to the trouble of splitting the $5 between the three he simply gives them $1 each and pockets the remaining $2 for himself.

Now, each of the men effectively paid $4, the total paid is therefore $12. Add the $2 in the waiters pocket and this comes to $14.....where has the other $1 gone from the original $15?
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10 dollar meal, 2 dollar tip. the three bucks went back to the three guys.

Originally Posted by jacobo

What if each of the men refused the dollar refund from the chef so they effectivly paid $4 each, the total paid is therefore $12 and the waiter then keeps the $3 + $2 in already in his pocket for a total of $17 of the original $15.
Here is one.
You're on a game show and the grand prize is a trip for two with your Fav DV member [trip desination your choice] and is behind one door of three doors. The other 2 doors are only lousy DV T-shirts .

The gameshow host asks you to select a door

Door 1 Door 2 Door 3

You pick door #1( 1,2,3 does't matter it stays closed). The host then opens one of the other two remaining doors to reveal one styling DV T-shirt. He now procedes to ask you, Would you like to change the door that you picked? (obviously you want to get the trip with a choice DV member, not the T-shirt so now question time...)

Do you switch doors or not, please provide the reasoning for your answer.
And yes, there is a correct answer.

BTW I hate this one......
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I too hate the logic behind this one. The logical thing to do would be to choose the other door because there is not a 50/50 shot because you are now only left with two choices as opposed to a 33/33/33 shot. But personally id stick with the same door because you still dont knwo which one is right.


ha, didnt realise somone got that already, just had to get rid of the mish mosh.

total cost of which is $15. They each contribute $5. The waiter gives them $1 each and pockets the remaining $2 for himself.


Basicly you have to find the 10 bucks. the cost of the meal.


total cost of which is $15. They each contribute $5. The waiter gives them $1 each($3) and pockets the remaining $2 for himself.

15-3-2=10

You only get one weigh? I don't need any. I trust my motor functions enough that I can tell if something is half as heavy as something else.