|
12 coins:
+ Show Spoiler +I can't see this one working. The problem is reduced to 8 coins of which 1 is counterfeit, with 4 real coins to help us and 2 weighings remaining. Unless I am missing something, here is how I would proceed to find the counterfeit:
Weigh 4 of the suspected coins against the 4 real coins (or two suspected coins on each side). Regardless of the result we're now down to 4 possible counterfeits and 1 weighing. Like the "hint" that was posted earlier says, this requires 2 weighings to solve from here.
I'd really like an explanation for this one...
48/2(9+3):
+ Show Spoiler +With the way I was taught math: 2. This is more of an argument starter than an actual math problem since a mathematician would have the good sense to include another parenthesis.
|
I guess I'll pick this one you label as your favorite.
5.) Suppose a rectangle can be (in some way) entirely covered by 25 disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer. Note: overlaps allowed of course.
+ Show Spoiler +I don't see the point in them being disks, but anyways lets convert them to squares since circles are dumb and give the same results. Radius of 1 so they are 2x2 squares (4 units squared is their area). It's completely covered, so it must have an area of 4*25 = 100 units squared.
Now we got new smaller disks with a radius of 1/2 meaning they are equivalently 1x1 squares. So the area of 100 1x1 squares is 100 units squared. Same as last time.
So the answer is an obvious yes, 100*1 is indeed the same as 25*4.
My problem with these kind of math/brain teaser questions is they normally rely on you having to assume things because if they actually give you all the relevant information like in #5, they are too incredibly easy.
|
26. suppose you have a chess board with 2 opposite corners cut out as in picture
![[image loading]](http://www.cl.cam.ac.uk/~mj201/images/mutilated-chess-board.gif)
there would be 62 squares in this cut out board. you have a set of domino pieces, each piece can cover exactly 2 adjacent squares of the chess board. Is it possible to cover (tile) the cut out chess board with exactly 31 pieces of dominos? if yes, how? if not, why not?
|
On April 10 2011 13:01 Befree wrote:I guess I'll pick this one you label as your favorite. 5.) Suppose a rectangle can be (in some way) entirely covered by 25 disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer. Note: overlaps allowed of course. + Show Spoiler +I don't see the point in them being disks, but anyways lets convert them to squares since circles are dumb and give the same results. Radius of 1 so they are 2x2 squares (4 units squared is their area). It's completely covered, so it must have an area of 4*25 = 100 units squared.
Now we got new smaller disks with a radius of 1/2 meaning they are equivalently 1x1 squares. So the area of 100 1x1 squares is 100 units squared. Same as last time.
So the answer is an obvious yes, 100*1 is indeed the same as 25*4.
My problem with these kind of math/brain teaser questions is they normally rely on you having to assume things because if they actually give you all the relevant information like in #5, they are too incredibly easy.
As I said:
On April 10 2011 00:46 stepover12 wrote:
Ahh, for number 5, showing that the area is enough to cover may suggest the answer is yes. But it is not a definite proof that there is a way to cover the rectangle with those smaller disks.
There is an elegant solution without having to make any other assumption. Hint:
+ Show Spoiler +
solution for 5 (challenge yourself, think about the hint before looking down below):
+ Show Spoiler +divide the rectangle into 4 smaller rectangles similar to the original. Each of these small rectangle can be covered by 25 small disks (of radius 1/2) the same way that 25 disks of radius 1 can cover the big rectangle. Putting them together, we get 100 small disks can cover the whole big rectangle. The answer is proven to be yes.
|
On April 10 2011 12:54 eekon wrote:12 coins: + Show Spoiler +I can't see this one working. The problem is reduced to 8 coins of which 1 is counterfeit, with 4 real coins to help us and 2 weighings remaining. Unless I am missing something, here is how I would proceed to find the counterfeit:
Weigh 4 of the suspected coins against the 4 real coins (or two suspected coins on each side). Regardless of the result we're now down to 4 possible counterfeits and 1 weighing. Like the "hint" that was posted earlier says, this requires 2 weighings to solve from here.
I'd really like an explanation for this one... 48/2(9+3): + Show Spoiler +With the way I was taught math: 2. This is more of an argument starter than an actual math problem since a mathematician would have the good sense to include another parenthesis.
A hint on the 12 coins
+ Show Spoiler +You don't actually have 8 coins that contain a counterfeit with 2 weighings left. You have 12 coins left, and you know which 4 (minimum, assuming the first weighing is unbalanced) or 8 (maximum, assuming the first weighing is balanced) are normal. You can use this to force certain balance. Another hint in the nested spoiler: + Show Spoiler +
You never make a weighing of 1 coin against another. You always make a minimum weighing of 2 coins.
As I've heard it before, I'm not going to give the full solution.
Edit: My guess on 26.
+ Show Spoiler +
I'm pretty sure it is impossible. You must use shapes of 1x2 to cover a board that can be subdivided into 3 smaller grids of 1x7, 1x7, and 6x8. No matter how you place the tiles, you will eventually be in a situation where you must cover a grid of 1x7 squares with 1x2 tiles. This will prove to be impossible.
|
hint for people solving the 12 coin problem: drawing a table of probability will help a ton since the answer might require you to mix up the coin to predict the out come a little bit. Anyone who could solve this in their head must either hv heard of this b4, google it or is a freaking genius.
it took me a full night back in grade 9th just to solve that problem with a whole bunch of logic math books surrounding.
|
On April 10 2011 13:09 Aequos wrote: A hint on the 12 coins + Show Spoiler +You don't actually have 8 coins that contain a counterfeit with 2 weighings left. You have 12 coins left, and you know which 4 (minimum, assuming the first weighing is unbalanced) or 8 (maximum, assuming the first weighing is balanced) are normal. You can use this to force certain balance. Another hint in the nested spoiler: + Show Spoiler +
You never make a weighing of 1 coin against another. You always make a minimum weighing of 2 coins.
As I've heard it before, I'm not going to give the full solution.
+ Show Spoiler +Since a hint was posted earlier covering the easier scenario where the initial weighing was balanced, thus reducing the possible counterfeits to 4, I'm assuming the rest of the problem is, indeed, 8 possible counterfeits with 2 weighings remaining. Are you saying I'm already in the wrong at this point?
For NB: The thought of bringing probability into a logic riddle is... worrying!
|
On April 10 2011 09:22 GertHeart wrote:10.) Easier than you think I had this in a leadership class. + Show Spoiler +
You will not be able to see the hat on your head.
Is the statement you should focus on, if you are smart enough you'd just take the hat off your head so you can see it.
These riddles usually go with something so obvious that you feel a little wetarded when you have to solve it.
Reminds me on "How fast can everyone touch the ball trick?" Our class of about 30 students all touched it in about 1 seconds flat after we figured it out, first we were passing the ball to each other got that down to about 20 something seconds, then we tried getting closer, got it down more, then we tried just handing it off in a circle got it down faster. Then we all put a finger on the ball and let go at the same time.
It just shows there is more than one solution, but which is best? In other words a lot of the solutions weren't wrong, and a lot weren't right, the ones where you have to sacrifice someone is just silly =P Nobody needed to die! just take the hat off and put it on your hand and say the color
No, there is no trick. These are not stupid scenarios with an easy trick that makes you say "Oh, I should have thought of that." These are legit problems that actually require effort to solve. The answer to 10 is just modulo arithmetic.
The 12 coins problem solution is rather convoluted if I recall correctly, and it actually takes a while to sit down and write it out, or maybe my solution was not quite as elegant.
EDIT: I think this is the answer + Show Spoiler +
Separate the balls into 3 categories 4 A's, 4 B's, 4 C's. Weight AAAA vs BBBB. If they are equal, then the solution is simple.
Say that AAAA> BBBB. Then we can weigh ABC vs BAB, and call this group 1 vs group 2. If ABC< BAB, then either B1 is lighter or A2 is heavier, and the answer follows.
If they are equal, then either the two A's that weren't weighed are heavier, or the 1 B that wasn't weighted is lighter.
If ABC>BAB, then either A1 is heavier, or either of the B2's are lighter, which is similar to the case above. In both cases we have 2 balls that are lighter or heavier, and one ball that is the opposite. Without loss of generality say AA are heavier, or B is lighter, and the rest of the balls are normal, denoted N. Then we can weigh AB vs NN. If AB is lighter, B is counterfeit. If AB is heavier, then that A is counterfeit. If AB is equal weight, then the one A that was not weighed is the counterfeit.
|
26. Chessboard
+ Show Spoiler +
There is no way to cover the chess board with dominos. Each domino will cover one black and one white space, but the corners removed were both black. This has left 32 white spaces and 30 black spaces, so at best there will be two white spaces remaining after using 30 dominos.
|
On April 10 2011 13:10 NB wrote:
hint for people solving the 12 coin problem: drawing a table of probability will help a ton since the answer might require you to mix up the coin to predict the out come a little bit. Anyone who could solve this in their head must either hv heard of this b4, google it or is a freaking genius.
it took me a full night back in grade 9th just to solve that problem with a whole bunch of logic math books surrounding.
I managed to solve the problem in my head but it took me a 2-3 hours doing it with a friend.
Ill give an additional hint u have three things that a balance tells u. Which sides weights more, how many coins are on each side and if u switch side u can see which coins have an effect and which have none.
|
On April 10 2011 11:58 Darkren wrote:Show nested quote +On April 10 2011 11:35 Everhate wrote:On April 10 2011 10:43 Darkren wrote:On April 10 2011 10:36 Everhate wrote:On April 10 2011 09:41 Frigo wrote:On April 10 2011 08:43 Darkren wrote:
Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible? I believe this approach will work...unless someone can point out the flaw: + Show Spoiler +
Place 3 coins on each side, set the rest aside
-> Whether they balance or not, this leaves you with 6 coins with 2 weighs remaining, as well as 6 coins that you know are real, thus the proper weight.
Of the 6 unknown, place 2 on each side of the scale
-> Will leave you with 2 coins, with 1 remaining, and 10 that you know are the proper weight.
Place one of the remaining coins on one side, with one of the known coins on the other.
-> Either the unknown coin will be heavier/lighter, or will balance, meaning the last remaining coin is the counterfeit.
This does not work u are supposing that on the second balance u take 2 on 2 that u will take the bad coin. If it balances out u are screwed. Read one of my posts above and look at the start of the problem and continue from there K, think I have it this time, heh, realized my earlier mistake and tried to ninja edit it right as you quoted :p + Show Spoiler +
Divide 4/4/4 and place them onto the scale, if they balance it's quite easy as you posted earlier, so we'll address if it does not balance. I'll label the side that's lighter as A and the side that's heavier as B for ease, with those known as K.
Set up the scales for 2A 1B/2A 1B. If they balance, weigh the 2 Bs that are not on the scales against each other, the heavier is the incorrect weight. If they do not balance, weigh the 2 As from the lighter side against each other.
If these 2 balance, the B was heavy. If they do not balance, the lighter one is the incorrect weight.
Ur close but not quite there + Show Spoiler +On your first part if they balance and u weight the 2 B together how do u know the heavier one is the incorect one? U don't know if it is lighter or heavier so it could be any of those 2  U also do not know which As to weight eachother for the same reason as above
+ Show Spoiler +As the B coins were those that were heavier in the first weighing, that means that either one of the 4 A coins are light or one of the 4 B coins are heavy. By splitting them as I did, all of the possible 'light' coins were on the scale, while the 2 B that were considered if those balanced could only be unbalanced if one was heavier than it should be, else the initial weighing would have been different (heavy/light split based on observation from the first unbalanced weighing)
Edit for clarification:
That is to say, if I arranged them using L for possible light, and H for possible heavy, the scale for the second weighing would look like LLH/LLH. If the left side were lighter, that would mean either one of the 2 Ls on the left was actually light, or the one H on the right was heavy, obviously the reverse is true if the right side registered lighter...bit difficult to express this in written form
|
On April 10 2011 15:20 Everhate wrote:Show nested quote +On April 10 2011 11:58 Darkren wrote:On April 10 2011 11:35 Everhate wrote:On April 10 2011 10:43 Darkren wrote:On April 10 2011 10:36 Everhate wrote:On April 10 2011 09:41 Frigo wrote:On April 10 2011 08:43 Darkren wrote:
Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible? I believe this approach will work...unless someone can point out the flaw: + Show Spoiler +
Place 3 coins on each side, set the rest aside
-> Whether they balance or not, this leaves you with 6 coins with 2 weighs remaining, as well as 6 coins that you know are real, thus the proper weight.
Of the 6 unknown, place 2 on each side of the scale
-> Will leave you with 2 coins, with 1 remaining, and 10 that you know are the proper weight.
Place one of the remaining coins on one side, with one of the known coins on the other.
-> Either the unknown coin will be heavier/lighter, or will balance, meaning the last remaining coin is the counterfeit.
This does not work u are supposing that on the second balance u take 2 on 2 that u will take the bad coin. If it balances out u are screwed. Read one of my posts above and look at the start of the problem and continue from there K, think I have it this time, heh, realized my earlier mistake and tried to ninja edit it right as you quoted :p + Show Spoiler +
Divide 4/4/4 and place them onto the scale, if they balance it's quite easy as you posted earlier, so we'll address if it does not balance. I'll label the side that's lighter as A and the side that's heavier as B for ease, with those known as K.
Set up the scales for 2A 1B/2A 1B. If they balance, weigh the 2 Bs that are not on the scales against each other, the heavier is the incorrect weight. If they do not balance, weigh the 2 As from the lighter side against each other.
If these 2 balance, the B was heavy. If they do not balance, the lighter one is the incorrect weight.
Ur close but not quite there + Show Spoiler +On your first part if they balance and u weight the 2 B together how do u know the heavier one is the incorect one? U don't know if it is lighter or heavier so it could be any of those 2 U also do not know which As to weight eachother for the same reason as above + Show Spoiler +As the B coins were those that were heavier in the first weighing, that means that either one of the 4 A coins are light or one of the 4 B coins are heavy. By splitting them as I did, all of the possible 'light' coins were on the scale, while the 2 B that were considered if those balanced could only be unbalanced if one was heavier than it should be, else the initial weighing would have been different (heavy/light split based on observation from the first unbalanced weighing)
+ Show Spoiler +Thats the whole problem u don't know that the B coins where heavier in the first weigting it could have been that or it could have been that the A coins were lighter
|
On April 10 2011 15:27 Darkren wrote:Show nested quote +On April 10 2011 15:20 Everhate wrote:On April 10 2011 11:58 Darkren wrote:On April 10 2011 11:35 Everhate wrote:On April 10 2011 10:43 Darkren wrote:On April 10 2011 10:36 Everhate wrote:On April 10 2011 09:41 Frigo wrote:On April 10 2011 08:43 Darkren wrote:
Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible? I believe this approach will work...unless someone can point out the flaw: + Show Spoiler +
Place 3 coins on each side, set the rest aside
-> Whether they balance or not, this leaves you with 6 coins with 2 weighs remaining, as well as 6 coins that you know are real, thus the proper weight.
Of the 6 unknown, place 2 on each side of the scale
-> Will leave you with 2 coins, with 1 remaining, and 10 that you know are the proper weight.
Place one of the remaining coins on one side, with one of the known coins on the other.
-> Either the unknown coin will be heavier/lighter, or will balance, meaning the last remaining coin is the counterfeit.
This does not work u are supposing that on the second balance u take 2 on 2 that u will take the bad coin. If it balances out u are screwed. Read one of my posts above and look at the start of the problem and continue from there K, think I have it this time, heh, realized my earlier mistake and tried to ninja edit it right as you quoted :p + Show Spoiler +
Divide 4/4/4 and place them onto the scale, if they balance it's quite easy as you posted earlier, so we'll address if it does not balance. I'll label the side that's lighter as A and the side that's heavier as B for ease, with those known as K.
Set up the scales for 2A 1B/2A 1B. If they balance, weigh the 2 Bs that are not on the scales against each other, the heavier is the incorrect weight. If they do not balance, weigh the 2 As from the lighter side against each other.
If these 2 balance, the B was heavy. If they do not balance, the lighter one is the incorrect weight.
Ur close but not quite there + Show Spoiler +On your first part if they balance and u weight the 2 B together how do u know the heavier one is the incorect one? U don't know if it is lighter or heavier so it could be any of those 2 U also do not know which As to weight eachother for the same reason as above + Show Spoiler +As the B coins were those that were heavier in the first weighing, that means that either one of the 4 A coins are light or one of the 4 B coins are heavy. By splitting them as I did, all of the possible 'light' coins were on the scale, while the 2 B that were considered if those balanced could only be unbalanced if one was heavier than it should be, else the initial weighing would have been different (heavy/light split based on observation from the first unbalanced weighing) + Show Spoiler +Thats the whole problem u don't know that the B coins where heavier in the first weigting it could have been that or it could have been that the A coins were lighter
+ Show Spoiler +But I do know it has to be one or the other, as otherwise that's the reason for the scales to have been unbalanced (if the left side rises up while the right sinks, for example, then either one coin from the left, or A set of coins, is light, or one coin from the right, or B set of coins is heavy, but not both.
The direction of the imbalance in the first weighing indicates which coins could possibly be light, or which ones could possibly be heavy, as you couldn't have a 'B' coin that's lighter than it should be, else the scale would have tipped in the other direction in the first weighing. This result of that weighing is how I label them as A and B, and supports the conclusions I'm making in the next weighing.
Possible I may not be explaining myself adequately, or may not be understanding what you're saying, heh =)
Edit: Just realized I mistyped my 3rd weighing conclusion...If you weigh 2 A's against each other, and they balance, the B you conclude is heavy was the B on the opposite side...managed to not realize i didnt type that correctly, gg me ><...updating
|
After having looked this riddle up, a few notes...
+ Show Spoiler +The original question, as well as some of the hints given in this thread, were less than helpful. The real riddle demands that you determine if the fake coin is heavier or lighter, which actually makes the riddle easier. Nice try, Everhate. Your idea was creative, but it fails because you are only looking for the coin with the odd weight instead of... conducting a more thorough investigation!
|
On April 10 2011 15:52 eekon wrote:After having looked this riddle up, a few notes... + Show Spoiler +The original question, as well as some of the hints given in this thread, were less than helpful. The real riddle demands that you determine if the fake coin is heavier or lighter, which actually makes the riddle easier. Nice try, Everhate. Your idea was creative, but it fails because you are only looking for the coin with the odd weight instead of... conducting a more thorough investigation!
+ Show Spoiler +Unless I managed to overlook something in the process, I believe, in each of the conclusions, which coin is indicated will also tell whether it is heavier or lighter than the rest, based on which stack it came from. That was the point of splitting them after the first weighing into possibly heavy or possibly light. Which of those 2 stacks has the 'fake coin' should evidently be too heavy if from the possibly heavy pile, or too light if from the possibly light pile.
That is to say, assuming that I'm left with 8 possibilities after using the scale for the first time, I know that in one set of 4, one of the coins may be lighter OR one of the coins from the other set may be heavier. One of these 2 conditions must be true for the results of the first 4/4/4 split to have been imbalanced, and the final determination of which coin is 'fake' will also determine whether it is heavy or light, depending on which stack it came from.
It's a slightly different approach than the other solution that was posted in the thread, but still believe it handles all outcomes.
|
Like any puzzle games, theres always an answers tab that we can check. Put the answers just like a result tab of a live report thread for people with low iq like me who cant even answer a single question after trying so hard.
|
Unless I managed to overlook something in the process...
+ Show Spoiler +I'm running your system through a flash simulator, and I think I've covered all the possibilities. Can't get it to fail...
|
On April 10 2011 02:01 Keniji wrote:Show nested quote +On April 10 2011 01:53 funk100 wrote:
I have one
this there is this village beside a mountin where all justice is decided by the accused picking one ball out of a bag with two in - one white one black- they then show the ball to the audience and justice is administerd. if the criminal picks the black ball they are thrown off the cliff (where the trails take place) if they pick the white ball they can walk free.
let us suppose that there is a man who has commited 10 murders but has continually, through chance, always picked the white ball and allways walked free. On the night of his 11th murder the brother of the murdered gets pissed as he wants justice, so he switches the bag so both the balls are black and so, he reasons he can get revenge. the murderer hears of this from the brother as he put his plan as his fb status and the murderer is a mutuall friend of him.
On the day the murderer manages to walk free from the cliff after the village agree he picked a white ball, how?
REMEMBER/rules
1) there are 2 black balls in the bag and they are not changed
2) he does not have a trick ball anywhere - he chooses a ball from the bag
3) the ball he chooses is the one he is represented by
4)where he is + Show Spoiler +just to clarify:
the ball he picks and actual show to the audience is also the ball he is represented by?
the ball the audience is convinced he picked first is the one he is represented by
|
|
|
On April 10 2011 18:21 Robstickle wrote:+ Show Spoiler +He rolls the ball in snow before showing it to the villagers?
in this country there is no snow, or any white things. the white balls are imported from a million miles away.
|
|
|
|
|
|
EPT
