Riddles

I (the counselor) could send 2 groups of 4 down 2 of the different paths and go on the third by myself. Then there would be at least two in each group that don't lie.

Good thought, but if you didn't find it and one group of four didn't find it, and the other group of four came back split two and two, again you wouldn't know whether that path was right or the unexplored one was. ;)

I just reread the riddle. If I come to a four-way intersection I have to have come from one way so that leaves only three unexplored paths. All eight of the campers could go on one path, I could go on the second and if no one gets to the campsite we know it is the third.

Hehe! Smarty! :P But actually, no, there are four different directions the campsite could be at. It's as if you came out of the woods and happened upon that intersection. :)

Pray you'll chose the right one and go :P jk…this seems to be a tricky riddle, but there's probably a catch. :P

Group #1 could have five people, group #2 could have three and I would be group #3.
Assuming both liars will definitely lie, if two people in group #1 say one thing but the other three say something else, the three would be telling the truth because there are only two liars. Based on how many liars are in group #1 you could figure out who in group #2 is telling the truth. If two in group #1 lie then all three in group #2 are honest. If one in group #1 lies that leaves one liar in group #2, so what two campers in group #2 agree on would be the truth. If no one in group #1 lies there are two liars in group #2, so the camper that says something different than the other two is honest.

Very well thought out! :) But, sigh… The only problem is that, like you said, it relies on the assumption that both liars will lie all the time, which is not necessarily the case. If the first group agrees that they did not find it and the second group is split two-to-one, again, that leaves two possibilities for where the campsite is. You're getting closer though! Hint: Try splitting up the group of five.

Okay. So I could have four groups, group #1 would have three campers, group #2 would also have three, group #3 would have two, and I would be group #4. If group #1 found the camp, but both liars were in that group, both liars could say that they didn't find the camp and the honest camper would say they did. Or maybe only one liar actually lied and two of the campers say that they found it but one liar says they didn't. Either way, because the group disagrees we would know there was a liar in that group. All the other groups agree that the camp is not on their path, so we would take the path group #1 took. If there is one liar in group #1 and one liar in group #2 we would believe whatever two campers in each group agreed on, because there has to be a liar in both groups since the whole group didn't agree, but there are only two liars. If group #3 finds the camp but there is a liar in group #1 and group #3 you would believe what two agreed on in group #1 ( that they didn't find the camp ) and you would take the path group #3 took. If group #1 finds the camp but both liars are in group #3, even if both campers in group #3 say they found the camp, you know group #3 lied not group #1 because there are three campers in group #1.

Yay! Correct! :D Here's a more precise way to handle the responses:

If the counselor finds the campsite, he leads his campers toward it. If there is disagreement within the two three-camper groups, believe the majority opinion (there can only be one liar per group). If there is disagreement within one group of three and one group of two, take the majority opinion of the three and ignore the two. If there is disagreement within only one group altogether, ignore the responses of that group. If there is no disagreement within any of the groups, ignore the group with two campers.

This method will always give a satisfactory answer about at least three of the four paths. Congratulations! :)

So….we're in the woods. its dark…. something wrong with that?
I'd put them all to work setting up a camp right then and there. "Sleep well boys, we got a lot of hiking to do tomorrow. And good thing about not eating dinner is the bears won't smell any dinner in us. we'll be Ok."

Researchers for the Massachusetts Turnpike Authority found over 200 dead crows near greater Boston recently, and there was concern that they may have died from Avian Flu.

A Bird Pathologist examined the remains of all the crows, and, to everyone's relief, confirmed the problem was definitely NOT Avian Flu. The cause of death appeared to be vehicular impacts.

However, during the detailed analysis it was noted that varying colors of paints appeared on the bird's beaks and claws. By analyzing these paint residues it was determined that 98% of the crows had been killed by impact with trucks, while only 2% were killed by an impact with a car.

MTA then hired an Ornithological Behaviorist to determine if there was a cause for the disproportionate percentages of truck kills versus car kills. He very quickly concluded the cause: When crows eat road kill, they always have a look-out crow in a nearby tree to warn of impending danger.

They discovered that while all the lookout crows could shout "Cah", not a single crow could shout "Truck."

Lol! xD But wouldn't this be better suited for the jokes thread?

Yes, but I couldn't find it. :P

LOL.

https://www.memverse.com/forums/7/topics/jokes :P

Here is a variant of the camper's problem riddle: Suppose there are five liars. In this case, what is the minimum number of campers needed to successfully solve the problem? (It should be under 20.)

So....we're in the woods. its dark.... something wrong with that? I'd put them all to work setting up a camp right then and there. "Sleep well boys, we got a lot of hiking to do tomorrow. And good thing about not eating dinner is the bears won't smell any dinner in us. we'll be Ok."

You have a very good point. Unfortunately, riddles like these (unless it's a trick problem) don't often allow room for alternate ways of thinking and "cheater" solutions to get out of the problem. xD

Here is a variant of the camper's problem riddle: Suppose there are five liars. In this case, what is the minimum number of campers needed to successfully solve the problem? (It should be under 20.)

18 total people? (17 campers plus the counselor)

Exactly! :) With the counselor taking path A, six campers each on path B and C and five campers on path D, a similar solution to the problem with two liars always gives an answer about at least three of the four paths.

Are y'all tired of this problem yet? :P Here's one more twist.

It turns out that eight campers is the minimum with two liars and time for only one search. Suppose there are only four campers, two of which still sometimes lie. However, this time you have 100 minutes of daylight - time for two searches before making a final decision.

Here are two riddles. Well technically, they're math problems.

Four people are hiking in the jungle at night when they come to a bridge. The bridge is very narrow, and only two people can cross at a time. The group also has one torch that you need to cross the bridge because it is dark. Person A can cross the bridge in 1 minute, B in 2 minutes, C in 5 minutes, and D in 8 minutes. How do you get all the people across the bridge if the torch burns out in 15 minutes?

You want to cross a river in a boat and bring along a goat, a wolf, and a cabbage. The problem is that the wolf wants to eat the goat, and the goat wants to eat the cabbage. However, nothing will be eaten if you are present. You are the rower and your boat can only hold one item besides yourself. How do you get all three across?

The counselor takes path A and all four campers go on path B for the first search.
For the second search the counselor takes path B and all four campers go on path C. Since the counselor has taken path B he knows who of the campers were lying in the first search so if the campers disagree again he knows who is honest. If the campers don't disagree in the first search the counselor would take path D so that even if the campers disagree in the second search he knows what is at the end of three of the paths.

1. Goat and human go across (leave the wolf with the cabbage)
2. Return and bring cabbage over
3. Bring goat back, leave it and take wolf over
4. Return, take the goat over, and you're done.

@Zeke - that's correct.

Here's a harder problem: Each digit in a simple equation is represented by a letter. Each letter represents a different digit. So for example, 50 + 51 = 101 could be represented as AB + AC = BCB. What digit is represented by Y in the equation
SEND + MORE = MONEY? There is no code.

Here are two riddles. Well technically, they're math problems. Four people are hiking in the jungle at night when they come to a bridge. The bridge is very narrow, and only two people can cross at a time. The group also has one torch that you need to cross the bridge because it is dark. Person A can cross the bridge in 1 minute, B in 2 minutes, C in 5 minutes, and D in 8 minutes. How do you get all the people across the bridge if the torch burns out in 15 minutes? You want to cross a river in a boat and bring along a goat, a wolf, and a cabbage. The problem is that the wolf wants to eat the goat, and the goat wants to eat the cabbage. However, nothing will be eaten if you are present. You are the rower and your boat can only hold one item besides yourself. How do you get all three across?

For riddle number one person A and B go first. Person A holds the torch and stops halfway across the bridge but person B keeps going. When person B gets off person C goes across. When person C gets off person D starts. When person D gets halfway across (where person A is holding the torch) person A walks the rest of the way across with person D.

Sorry, I forgot another piece of information. When two people go across the bridge, they must go at the speed of the slowest person. So person A must travel at the same speed as person B.

I've heard that riddle before. I think I posted it on the blog quite a while back. I won't give it away though. xD

The counselor takes path A and all four campers go on path B for the first search. For the second search the counselor takes path B and all four campers go on path C. Since the counselor has taken path B he knows who of the campers were lying in the first search so if the campers disagree again he knows who is honest. If the campers don't disagree in the first search the counselor would take path D so that even if the campers disagree in the second search he knows what is at the end of three of the paths.

thinks for a minute

^You know when "the judges are deliberating" that you gave an answer they didn't expect but that is probably still right. xD^

Okay, I would say yes, that's correct! The only assumption it makes is that the counselor can conclusively tell the campers apart. That shouldn't be too hard. :P
Here's my answer, pretty similar, but it actually uses at most one camper in the second round (!): The counselor searches one path, say A, in the first round, while all four campers go down B. When they rendezvous, if the camper found the campsite at A, then it is A. If three or four campers agree that it is B, then it is B. If three or four campers agree that it isn't B, then the counselor searches C in the second round and all the campers rest.
Now for the hard case. If two say it's down B and two say it isn't, the counselor searches B again. One of those who said it wasn't B is sent down C. Now if the counselor finds it at B, then it is B. If he doesn't find it, then the camper sent down C must be telling the truth, since the liars were those who said it was down B.

@Zeke - that's correct. Here's a harder problem: Each digit in a simple equation is represented by a letter. Each letter represents a different digit. So for example, 50 + 51 = 101 could be represented as AB + AC = BCB. What digit is represented by Y in the equation SEND + MORE = MONEY? There is no code.

I'll get back to you on that one. :D (I've done similar riddles before x) )

This one is really hard. Tell me if it's too hard. xD

There are 13 logicians in a room, all wearing jackets. On the front of each logician is a name tag and all the logicians have different names. On the back of some of the jackets is a big X. Each of the logicians can see the back of everyone else's jacket, but not his own. Initially, someone comes into the room and says "At least one of you has an X on his back." The problem is for each logician to figure out whether he has an X or not.
They do this in the course of several rounds. In each round, the logicians who have not yet decided whether they have an X on their backs speak in alphabetical order. Each logician either says:

• I don't know whether I have an X on my back.
• I don't have an X on my back.
• I do have an X on my back and at least one other logician does also but has not yet said that he does.
• I do have an X on my back and all other logicians who do have already said so.

They are not allowed to say anything else. As soon as a logician decides, that is, announces, that he does or doesn't have an X on his back, he stops speaking.

The result is this: In the first round, four people decide. In the second round, three people decide. One decider in the second round says that there are more X's. In the third round, the remaining six decide.

With these results, which logicians have X's on their backs?

@Zeke - that's correct. Here's a harder problem: Each digit in a simple equation is represented by a letter. Each letter represents a different digit. So for example, 50 + 51 = 101 could be represented as AB + AC = BCB. What digit is represented by Y in the equation SEND + MORE = MONEY? There is no code.

Y = 2. That was a good workout for my brain, especially when I initially did it all in my head! :P

It might be helpful to put the problem in column format:

@..S E N D@
@.+M O R E@
_______
@M O N E Y@

1. Assuming that M is non-zero (because after all, it is the beginning of two numbers!) S + M must equal a two-digit number with a 1 in the tens place (even if carrying is allowed); thus, M = 1.
2. Depending on whether carrying happened from the hundreds column, O can be either 0 or 1 if M is 1. But O cannot be 1 since M is already established to be 1, so O = 0.
3. We observe then that carrying must have occured in the tens column: If O = 0 and no carrying occured, then E must equal N, but that is impossible. So, E + 1 = N. (Back to this in a little bit.)
4. At this point S can be either 8 or 9, depending on whether carrying occured in the hundreds column. If carrying did occur, S = 8, E = 9 and N = 0 - but that is impossible, O is already 0. Thus carrying did not occur in the hundreds column, and S = 9.

Here is everything that is known so far:

~1……..1…?~
@..9 E N D@
@.+1 0 R E@
_______
@1 0 N E Y@

1. Now observe the tens column. Since there is a 1 carried to the hundreds column, N + R ≥ 9 (not 10, because we don't yet know whether carrying occured in the ones column). Specifically, exactly one of these equations is true: N + R = E + 10 or N + R + 1 ^(for carrying from the ones column)^ = E + 10. Substituting (E + 1) for N (established in #3) in these equations and simplifying yields R = 9 or R = 8. But S = 9, so R must equal 8. By induction, we also conclude that carrying must have occured in the ones column.
2. Since carrying occured in the ones column, D + E ≥ 10. If D + E = 10, Y = 0 - but O already equals 0. Similiarly we can rule out D + E = 11. If D + E = 12, the two letters must be 5 and 7 in some order (all other combinations are ruled out by the same presupposition that no two letters represent the same digit). If D + E = 13, then they must be 6 and 7. Any larger sum of D and E is also ruled out (check it).
3. Now if D + E = 13, either {D = 6 and E = 7} or {E = 6 and D = 7}. The former cannot be true because it would mean N = 8 (see #3), but R is already 8. Neither can the latter be true because it would make both D and N equal 7. So this possibility is ruled out; D + E must equal 12. This means Y = 2.
4. D + E = 12, and either {D = 5 and E = 7} or {E = 5 and D = 7}. The first possibility is not true because again, it would require N to be 8. So E = 5, D = 7, and E + 1 = N = 6.

Here is the finished problem:

~1……..1…1~
@..9 5 6 7@
@.+1 0 8 5@
_______
@1 0 6 5 2@

A tribe has grown a wonderful harvest of crops. The chief is very smart and has allocated all the crops to three families of different sizes. To each family, he gives 10 tons more than half of what remains after he has given to the previous family. How many tons of crops did the chief start with?

140 tons, I believe. The first family got 80 tons (140 / 2 + 10), the second 40 tons ((140 - 80) / 2 + 10), the third 20 tons ((140 - 80 - 40) / 2 + 10).
Great riddle. :) Maybe my little brother could tell you how long I sat around furiously punching numbers into my calculator.

My favorite paradox (and also the shortest one I know):

This statement is false.

@Christian - You know that riddle you posted about the men with red and blue hats and having to guess them? Well, I was on Khan Academy the other day and they had a whole video about that, and it was like 15 minutes long. After I watched it, it made so much sense, and it seemed so simple! They have hardly any videos on there about brain teasers, but I thought it was neat they had that one. =p And they had a couple other cool ones as well. =)

My favorite paradox (and also the shortest one I know): This statement is false.

That is truly false.

Yes, that's right! Here is a riddle for your little brother. :)

If the number of minutes InSoloChristo sat punching numbers into the calculator equals 1/4 of the number of papers I used to figure out Matthew's riddle, how many pieces of paper did I use? Lol

My favorite paradox (and also the shortest one I know): This statement is false.

That is truly false.

Agreed. But, to the amateur logician, it seems contradictory to say that a statement is false when the statement itself states that it is false. That makes it seem like the statement is true. But, just because a statement is false doesn't mean the opposite of the statement is true. ^I think. xD^

This one is really hard. Tell me if it's too hard. xD There are 13 logicians in a room, all wearing jackets. On the front of each logician is a name tag and all the logicians have different names. On the back of some of the jackets is a big X. Each of the logicians can see the back of everyone else's jacket, but not his own. Initially, someone comes into the room and says "At least one of you has an X on his back." The problem is for each logician to figure out whether he has an X or not. They do this in the course of several rounds. In each round, the logicians who have not yet decided whether they have an X on their backs speak in alphabetical order. Each logician either says: * I don't know whether I have an X on my back. * I don't have an X on my back. * I do have an X on my back and at least one other logician does also but has not yet said that he does. * I do have an X on my back and all other logicians who do have already said so. They are not allowed to say anything else. As soon as a logician decides, that is, announces, that he does or doesn't have an X on his back, he stops speaking. The result is this: In the first round, four people decide. In the second round, three people decide. One decider in the second round says that there are more X's. In the third round, the remaining six decide. With these results, which logicians have X's on their backs?

Here's a hint for this riddle: What if there is only one logician with an X on his back? He would then see no X's, and since he knows that there is at least one X, he must be the one. What if there are two logicians with X's? The first with an X would have no basis for deciding, seeing one other X besides his own, but what about the second? Can the uncertainty of one logician prompt the reasoning of another?

If that one's too hard, here's another one slightly easier. I think it might have been posted on here before. xD

There are 72 prisoners kept together in a prison with eight cells, nine prisoners in each cell. The cells form a square, with a courtyard in the middle where they are allowed to congregate every day for meals. The prisoners are also allowed to move freely between the cells. Each day, however, the jailer counts the number of prisoners on each side of the prison to make sure none have escaped. Since the prisoners are allowed to move around, he does not particularly worry about the number of prisoners in each individual cell, as long as the total on each side of the prison is 27. (Is there a problem with this method of counting the prisoners? xD)

One day the prisoners find an escape route in which up to four prisoners can escape during the night. (Assume that all 72 prisoners are cooperative with each other and are interested in the common good, per se.) They begin escaping, but those remaining in the prison cleverly arrange themselves in the cells so that when the jailer counts, he will still count 27 on each side of the prison.

What is the maximum number of prisoners that can escape in this way without arousing the (admittedly naive) jailer's suspicions?

In round #1 three logicians say that they don't have an X on their back and one says that he does and at least one other logician does and hasn't said so yet. If two more logicians have an X on their back, and speak after the logician in the first round who says someone else has an X, but say they don't know if they have an X or not, they would both know that they had an X on their back because they could see only one X but the other logician with the X they saw also saw an X. In round #2 the three logicians that decide go last. The first one that decides says he has an X and at least one more logician does and hasn't said so. The second says that he has an X and all other logicians who do have said so. The third says he doesn't have an X. In round three the last six say they don't have an X.

Is that right? Or close?

Yeah, you're close! There are indeed exactly three logicians with X's! One caveat, though: How does the first logician with an X know that he has one?
One thing I may have forgotten to emphasize is that the logicians always go in alphabetical order in each round. You can number them logicians 1 to 13 or give them names beginning with A through M if you want. :P

I think these are the responses:

Round 1 ( If there is an X before the number, that logician has an X on his back.)

1- I don't have an X.
2- I don't have an X.
3- I don't have an X.
4- I don't have an X.
X5-IDK
X6-IDK
7- IDK
8- IDK
9- IDK
10- IDK
11- IDK
12- IDK
X13- I have an X and there are more X's

Number 13 would know that he has an X on his back because if only 5 and 6 had X's, when 5 said he didn't know,6 would know he had an X.

Round 2

X5- IDK
X6- I have an X and there are more X's.
7- IDK
8- IDK
9- IDK
10- IDK
11- IDK
12- IDK

Number 6 knows that 5 has to see an X since he said he didn't know, but the only X 6 sees in on 5.

Round 3

X5- I have an X and there are no more X's.

Then everyone else in round 3 decides that they don't have an X.

5 knows he has an X because in round 2, 6 says sees another X but 5 doesn't see it.

This riddle might be really simple or really difficult (or just confusing), depending on who you are. :P

A mathematician has a $1,000 debt, which he must pay on the next day. He also has a gem worth $1,000, but when he tries to sell it, potential buyers are deterred by the large number 1,000. He knows that in order to sell the gem, he will have to resort to the age-old marketing gimmick of pricing the gem at $999.99. However, if he does this, he will be $0.01 short of paying his debt, because the market he is at does not allow any prices to be rounded.
The mathematician comes up with a brilliant plan that will both attract buyers, and give him enough money to pay his debt. What price does he set for the gem?

Can he just give the gem to the person he is in debt to as payment?
Or it could be $999.99 plus sales tax.
Or he could advertise it at the low price of $999.99 ( just pay $.01 shipping and handling.)

There are no such complications, I'm afraid. I'm under the impression that he's trying to sell it himself at a place where people will come to buy such things, and the only money he'll get is from the buyer.
You never know; maybe he could pick up one penny from the street. :P But the answer has to do with the fact that he's a mathematician, and he knows how to manipulate numbers.

I'll give you a hint: while he can't round the price, he is not necessarily confined to the hundredth's place. He could go out to the thousandth's - or more - as long as he doesn't round.

Does he set the price at $1.0 x 10^3? :P

Good idea. :) It might look a bit unrealistic to buyers, though, as would $10³. The mathematician uses a slightly altered version of the basic gimmick, $999.99.
But since we're being perfectly honest, a relatively simple answer like yours might actually be more useful in real life than mine, which would require more explanation in a world where people have short attention spans and are highly critical of pretty much everything. (Such as… well, ours.) Oh well… :P

It's probably not this simple, but does he set the price at $999.999?

Closer, but he's still one tenth of a cent away from $1,000 unless he rounds, which he can't do. I also doubt any buyer would be thrilled about cutting a penny into ten pieces. :P (No cutting is necessary.)

Hold on… that gives me an idea. How about $999.99999……?
(It can be proven mathematically that 0.999999… to infinity equals exactly 1.)

Hold on... that gives me an idea. How about $999.99999......? (It can be proven mathematically that 0.999999... to infinity equals exactly 1.)

I think that's the answer :)
x = 0.9999999999…
=10x = 9.999999999…
= 10x -x = 9.999999999… - 0.9999999999…
= 9x = 9
= x = 1.