It's been awhile since i done this but I believe the answer is 1000111101.101110111
I cheated a bit though, I used the calculator to help me with the power calculation.
Ill wait to see if i'm right before I place a new test.
There are many different ways to represent binary, and my question was unclear because I couldn't think of the name for what I was asking for. What I was looking for was 1000111101.011. This representation continues the power series below the point, such that .1 is 1/2, .01 is 1/4, .001 is 1/8, etc... .375 is 1/4 + 1/8, so would be .011.
This is a good representation because the result can be normalized and raised to a power, as is done in the IEEE 754 standard. Does this count as TMI?
Anyway, I'll grant you the correct answer, since the number before the point is 573 and the one after is 375. Go ahead and ask another. :-)
Ok. This is a logics question.
There are three things on the shore. A fox a goose and a sack of grain. You want to get all three of these things to the other side of the shore without them eating eachother. You have a ship that can carry two things at a time across. How do you get all three across without having one of them eating another at any stage of the problem.
Take the fox and the grain accross and leave them on the other side, then go back and get the goose... (I am assuming that the fox is a carnivore here)...
You are on one side of a river. You have a fox, a goose, and a sack of grain. The fox cannot be left unsupervised with the goose, because he will eat the goose. The goose cannot be unsupervised with the grain, because he will eat the grain. In your boat you can only carry one across at a time. How do you get all three across to the other side of the river without any being eaten?
Take the goose across first. Leave it on the other side, come back and get the fox. Take the fox across, leave it on that side and take the goose back. Leave the goose behind, pick up the grain and take it over, leave it with the fox, and go back and get Mr goose. Eureaka!
Wait a second, are we dealing with a powerboat here? Cause if so, you can probably get away with leaving the goose and grain on one side as long as you don't stop to do a little fishing. Get back before the goose has been able to eat much of the grain (5-10 minutes) and you should be fine.
OK, While we are applying mathematical logic to real life, here's one for you:
A Hunter rose early, ate breakfast, and headed South. Half a mile from camp he tripped and skinned his nose.He picked himself up, cursing, and continued South. Half a mile further along, he spotted a bear. drawing a bead, he pulled the trigger, but the safety was on. The bear saw him and headed East at top speed. Half a mile later the hunter caught up, fired, but only wounded the beast, which limped on towards the east. The hunter followed and half a mile later caught and killed the bear. Pleased, the hunter walked the mile North back to his camp to find it had been ransacked by a second bear.
"What colour was the bear that tore up his camp?"
The clues are all there.
White. The hunter's camp must have been a little over a mile north of the south pole, because after walking a mile south, walking one mile east was far enough to go all the way "around he world" and return to where he started. Putting him just north of the south pole, where the circumference of the earth would only be one mile. Since he was in Antarctica then, the bear was a polar bear.
Here's an easy one:
3 boxes are labelled "Apples" , "Oranges" and "Apples & Oranges".
Each label is incorrect.
Label the boxes correctly after only selecting one piece of fruit, from one box.
(Stopwatch starts...now)
Man: How many birds and how many beasts do you have in your zoo?
Zookeeper: There are 30 heads and 100 feet.
Man: I can't tell from that!
Zookeeper: Oh, yes you can!
.....Can you?
OK.
Here's a more challenging one, but you should still get it quickly.
You have 12 identical looking coins, one of which is counterfeit.
The counterfeit coin is either heavier or lighter than the rest.
The only scale available is a simple balance.
Using the scale no more than 3 times, find the counterfeit coin.
(tick tick tick)
ok I solved your problem but it's a little tricky to explain so I am going to use a lettering system to do it:
Step 1)
Divide the twelve coins into 4 piles of 3.
Step 2)
Place one of the piles on each side of the balance. If the two piles are unequal, label each of them 'A' (making note of which is lighter), and label the other two piles 'B'. If the two piles are equal, label the other two piles 'A' (it is unknown which is lighter), and the two piles on the scale 'B'. Now the counterfeit is in one of the 'A' piles either way. Note: it is still unknown whether the counterfeit is heavier or lighter than the rest, whether the piles were equal or not.
Step 3)
Take one of the two 'A' piles (the lighter one if it is known), and one of the two 'B' piles and balance them. If they are equal, the counterfeit is in the 'A' pile not balanced (and the counterfeit is known to be heavier than the rest if the piles were unequal in Step 2). If they are unequal, the counterfeit is in the 'A' pile balanced (and the counterfeit is lighter than the others if the piles were unequal in Step 2. It is heavier than the others if the piles were equal in Step 2). Note: if the piles were equal in both Step 2 and Step 3, you don't know whether the counterfeit is heavier or lighter than the rest, but you still know which pile it is in.
Step 4)
Take the pile containing the coin, and balance one of its coins against another. If they are equal the counterfeit is the third coin, not balanced. If they are unequal, refer to your notes on whether the counterfeit was heavier or lighter than the rest, to determine whether the heavier or lighter coin on the balance is the counterfeit.
In the specific case where the piles are equal in both Steps 2 and 3, and the coins are unequal in Step 4, this problem cannot be solved using only 3 balances, at least not that I can think of. Also, if the piles are equal in both Steps 2 and 3, and the coins are equal in Step 4, you have identified the counterfeit, but still don't know if it is heavier or lighter than the other coins (although it was not necessary to determine that in this case).
I laid awake for hours last night as I couldn't get this puzzle out of my head. In every scenario I went through, there were always cases where you could end up with too many possibilities in the third step; If you divided them inot 4 groups of 3 there was a problem if the first three groups you compare are equal, as you would be left with three suspects but not know whether the counterfeit was lighter or heavier (as Torque pointed out above); You could also divide into three groups of 4, but then you have a problem if the first two groups you weigh are -not- equal; you can use the third group to determine which group the fake is in, and also if it is lighter or heavier, but you still have four suspect coins; even if you remove one, and replace it with a legit coin, and then weigh two of the suspect coins against the one suspect coin and the legit one, there is still a chance that the fake can be in the group containing two sus coins, and you still can't know which one it is!!! So the best I could do is that in either case, if you have very bad luck, you may still have two suspects left at the end.
put them into 6 groups of 2
place 2 on at a time until the balance unbalances
take one from the unbalanced group of 2 and compare it with another coin, if it balances, than its the other in the group of 2, if not, take your unbalanced coin and throw it in the garbage for causing so much grief
There is a solution that will work in every instance. Rather than give it immediately, I am going to give some clues as you are so close. (except for the garbage bin idea).
1) Number all the coins first.
2) This puzzle requires you to think about what can be deduced about the coins - both the ones you are weighing, and from that, about the ones you may elect NOT to weigh...(as Torque was doing)
3) Try starting with 3 groups of 4, (not 4 of 3).
Solution tomorrow...
Sorry but I have already spent far too much time on this on already!! Honestly I must have tossed and turned for at least an hour last night as my mind would not shut up!
Sam, apologies for your insomnia! You'll be pleased to know the solution is not a 'one liner', it is a process...Both you and Torque were so close in the last section of your figuring...
Solution:
Weigh coins 1,2,3,4 against 5,6,7,8.
***
If they balance, weigh coins 9 & 10 against coins 11 & 8 (we know from the first weighing that 8 is a good coin). If they balance, we know coin 12, the only unweighed one, is the counterfeit. The third weighing indicates whether it is heavy or light.
***
If however, at the second weighing (above), coins 11 & 8 are heavier than coins 9 & 10, either 11 is heavy; or 9 is light or 10 is light. Weigh 9 with 10. If they balance, 11 is heavy. If they don't balance, either 9 or 10 is light.
***
Now assume that at first weighing the side with coins 5,6,7 & 8 is heavier than the side with coins 1, 2, 3 and 4. This means that either 1,2,3, or 4 is light, or 5,6,7 or 8 is heavy. weigh 1,2 & 5 against 3, 6 & 9. If they balance, it means that either 7 or 8 is heavy or that 4 is light. By weighing 7 & 8 we obtain the answer, because if they balance then 4 has to be light. If 7 & 8 do not balance, then the heavier coin is the counterfeit.
***
If, when we weigh 1,2 & 5 against 3,6 & 9, the latter side is heavier, then either 6 is heavy, or 1 is light or 2 is light. By weighing 1 against 2 the solution is obtained.
***
If however, when we weigh 1,2 & 5 against 3,6 & 9 the latter side is lighter, then either 3 is light or 5 is heavy. By weighing 3 against a good coin the solution is easily arrived at.
Agreed Sam. I started doing this mix and match in my head yesterday but I think I would have needed to write it down as well. Or perhaps play with some numbered paper coins :)
Noobguy - as either 9 or 10 must be light, when weighed together at step 2, the lighter one is the counterfeit. You only need to go to the third weighing if they balance.
can you quess?