Logic Puzzles

[blackdiamond]

Ever since i was a child i have loved solving logic puzzles with my friends. I wish to use this opportunity to start a thread where people on this very intelligent forum can share the interesting puzzles they know with others who have an interest in this.

Before i share my first puzzle, here are my rules (i am God here, since i started this thread!):

1. When you solve a puzzle shared by someone, you must PM that person instead of giving the solution publicly, but you can discuss any interesting experiences or thoughts - about your mind, the brain, the puzzle, IQ, etc - that you realised during your process of solving that problem. The reason u should not put your solution on the forum is because there will be others who might want to attempt the same problem after you!

2. No boringly long puzzles should be shared here - the kinds that have many paragraphs and too many things to remember. Those are not interesting!

3. The puzzles do not have to be restricted to "mathematical" type; anything is welcome.

4. If you share a puzzle here and someone answers it correctly (by PM-ing you), you can share their names with us so we can 'admire' them - unless the list of those who have solved is too long (possibly meaning the puzzle was too simple, i think).

4. other rules will come from other people as we proceed; and we can then edit this first entry (of rules).

Okay, i will begin by sharing one of the most interesting puzzles i have ever solved. unfortunately, none of my friends here (in Zambia) has solved it yet, but they don't want me to give them the answer. Okay, here goes (or perhaps i should have started with a simpler one? no!):

PUZZLE 1: You have a balance scale and 12 coins, 1 of which is counterfeit. The counterfeit weighs less or more than the other coins. Can you determine the counterfeit in 3 weightings, and tell if it is heavier or lighter?

(Okay, so PM me the answer NOW if you can think of it. You can also post your own puzzles immediately - you don't have to wait until you have solved this one. If you don't know the answer to a puzzle, you can still bring it here.)

[fatdogs12]

PUZZLE 1: You have a balance scale and 12 coins, 1 of which is counterfeit. The counterfeit weighs less or more than the other coins. Can you determine the counterfeit in 3 weightings, and tell if it is heavier or lighter?

(Okay, so PM me the answer NOW if you can think of it. You can also post your own puzzles immediately - you don't have to wait until you have solved this one. If you don't know the answer to a puzzle, you can still bring it here.)

Umm I am guessing that all the other coins weigh the same or am I wrong?

[blackdiamond]

Umm I am guessing that all the other coins weigh the same or am I wrong?

Yes they do.

[jwoolcutt]

The following puzzle I got out of Doug West's "Fundamentals of Mathematics" book. It doesn't look like it has an answer, but it does. I will double check to make sure my wording is correct so there are NO TYPOS below and there is no "extra information" to be had.

Two mail carriers are on their route and happen to run into each other. Let's say their names are Bob and Tom.

Bob: "Tom it's been awhile since we met. I remember you had three kids but I forgot how old they are!"

Tom looks around, "Well Bob, the product of their ages is the number of windows on the building across the street."

Bob looks at the building, "Well, Tom, that doesn't tell me how old they are!"

Tom thinks for a minute, "You're right! The sum of their ages is your age Bob."

Bob thinks, and says, "You know, I STILL can't figure it out!"

Tom says, "Oh silly me how could I forget: My middle child has red hair!"

Bob says, "Oh thats right! Now I know how old they are!"

Assuming Bob is perfectly rational (i.e. doesn't make any arithmetic mistakes) how old are the children, and how does Bob know?

Enjoy!

Jacob Woolcutt -- Math Major

[Qwertz]

I think in this one it is important to know the product (in this case the number of windows). Given that, I believe I can determine the ages of the children and the age of Bob.

-Q

[jwoolcutt]

I think in this one it is important to know the product (in this case the number of windows). Given that, I believe I can determine the ages of the children and the age of Bob.

-Q

Indeed that would facilitate the solution, but it is NOT NECESSARY You do not require those numbers to solve the problem, though you _could_ make the reasonable assumption that the man was less that, say 200 years old or so.

Enjoy!

[Qwertz]

I still don't think it works. The children could be 11, 12, and 13, which makes Bob 36. But the building across the street had better be a massive glass skyscraper with 1716 windows. They could also be 2, 6, and 10, giving Bob an age of 18 (making the assumption that Tom is much older than him) and a more manageable 120 windows. Both men could be much older, and the children could be 20, 22, and 23, making Bob 65 and giving 10120 windows. There's still too much information missing.

PS - The remark about red hair serves only to tell us that there are no multiple-births involved. Tom would not have said "middle child" unless there were no twins or triplets among the three.

[aleph_0]

PUZZLE 1: You have a balance scale and 12 coins, 1 of which is counterfeit. The counterfeit weighs less or more than the other coins. Can you determine the counterfeit in 3 weightings, and tell if it is heavier or lighter?

Would the question "How do you determine the counterfeit and whether it is heavier or lighter in three weightings?" work just as well (i.e., is this a trick question)? And I assume you want both a complete and sound procedure (i.e. one that does not determine that all are the coins are counterfeit [which system would indeed tell you when you have the counterfeit but, unfortunately, lie to you when you do not have a counterfeit] and which will tell you whether you have the counterfeit in all possible situations, rather than just in--say--the situation where you randomly happen to choose the counterfeit).

[Qwertz]

This is the one I'm working on right now:

There are five houses, each of a different color, in a row. Each house is occupied by a person of a different nationality. Each person has a different pet, a different favorite beverage, and plants different flowers in his yard. The following rules apply:

1) The Englishman lives in the red house.

2) The Spaniard owns the dog.

3) The owner of the green house drinks coffee.

4) The Ukranian drinks tea.

5) The green house is to the right of the ivory house.

6) Geraniums are planted by the owner of the snails.

7) Roses are planted in front of the yellow house

8) The owner of the house in the middle drinks milk.

9) The Norwegian lives in the house on the far left.

10) The marigolds are next door to the fox.

11) The roses are next door to the horse.

12) The person who plants lillies drinks orange juice.

13) The Japanese grows gardenias.

14) The Norwegian lives next to the blue house.

Who owns the zebra, and who drinks water?

[Eric]

This is one of my favorites:

Mr. and Mrs. Jones went to a dinner party with three other married couples. At the party, various handshakes took place, but, of course, no one shook his or her own hand and no shook hands with his or her own spouse.

After the party, Mr. Jones asked each of the others how many people he or she had shaken hands with, and he was surprised because each person gave a different answer.

How many people did Mrs. Jones shake hands with?

- Eric

[Capitalism Forever]

I still don't think it works. The children could be 11, 12, and 13, which makes Bob 36. But the building across the street had better be a massive glass skyscraper with 1716 windows. They could also be 2, 6, and 10, giving Bob an age of 18 (making the assumption that Tom is much older than him) and a more manageable 120 windows. Both men could be much older, and the children could be 20, 22, and 23, making Bob 65 and giving 10120 windows. There's still too much information missing.

PS - The remark about red hair serves only to tell us that there are no multiple-births involved. Tom would not have said "middle child" unless there were no twins or triplets among the three.

It's you who are missing some information that's contained in the puzzle! Take a closer look.

[miseleigh]

The following puzzle I got out of Doug West's "Fundamentals of Mathematics" book.

This book doesn't seem to exist.... Do you have the title and author right? I've been trying to find the original question, and I can't- it doesn't appear that Doug West wrote a book called 'Fundamentals of Mathematics.'

[jwoolcutt]

I still don't think it works. The children could be 11, 12, and 13, which makes Bob 36. But the building across the street had better be a massive glass skyscraper with 1716 windows. They could also be 2, 6, and 10, giving Bob an age of 18 (making the assumption that Tom is much older than him) and a more manageable 120 windows. Both men could be much older, and the children could be 20, 22, and 23, making Bob 65 and giving 10120 windows. There's still too much information missing.

PS - The remark about red hair serves only to tell us that there are no multiple-births involved. Tom would not have said "middle child" unless there were no twins or triplets among the three.

As I said in the preface to the puzzle, it doesn't seem like there is a solution, but there actually is and it is solvable. There is no information missing and the solution is unique. One further hint: Place yourself in their shoes. If you were him you would know the number of windows and your own age, so WHY couldn't you figure it out even then?

Best,

Jacob

This book doesn't seem to exist.... Do you have the title and author right? I've been trying to find the original question, and I can't- it doesn't appear that Doug West wrote a book called 'Fundamentals of Mathematics.'

Whoops! You're absolutely right. I've got a bunch of these introductory math book lying around and I picked the wrong title from memory. In fact the problem was pulled from his book "Mathematical Thinking: Problem Solving and Proofs" This is especially embarassing--he was my undergraduate advisor! I apologize for the wrong citation.

That said, his book is a great way to start learning abstract mathematics, it has a ton of problems to solve and fantastic exposition. *shameless plug*

[Blinky]

There are four puzzles. They are nice, but rather easy (except of the fourth one) so don´t write my your answers.

1) There are five hats (two white and three black) and three men (each wearing one hat). They have their eyes covered. Then one of them has a look at the heads of the other two and says that he doesn´t know which colour does his hat have. Then the second one has his eyes uncovered and again he says that he doesn´t know what colour has his hat. After that the third man without uncovering his eyes says that he knows which colour has his hat. What colour is it?

2) Rule: If one card has "4" at one side it must have "E" at the other.

You have four cards. You see that they are: "4" , "7" , "E" , "G" . What card(s) do you need to turn up to prove that the rule isn´t false?

3) You can choose one of the doors. Behind one of them is beautiful girl (boy in the case of women). After you choose one of them, magician occures and says reveals that behind one of the remaining doors is nothing. Now you have a possibility to change your choice. Would you rather choose the same as in the first pick or change your choice to other door? Or do you think that it doesn´t matter?

4) There are two doors and each one is guarded by one soldier. Behind one of them something nice waits, the other one leads to the death (soldiers know what is hidden behind). One of the soldiers always lies and the other one always tells the truth. You have one question for one of them and you are supposed to choose the right door. What is the question?

I hope that the puzzles are clear. If they are not, just ask me and I will try to clarify them.

PUZZLE 1: You have a balance scale and 12 coins, 1 of which is counterfeit. The counterfeit weighs less or more than the other coins. Can you determine the counterfeit in 3 weightings, and tell if it is heavier or lighter?

I have known that one before and as far as I remember when I gave it to my friends after solving it myself they came with three different solutions. So if you want to have it even harder try to find them all (if there are any - I don´t want to solve it again and I remember just one of them).

[Qwertz]

Well I figured out some of the things I was missing, but I am still not coming up with only one possible solution. Here's as far as I got:

We are looking for two sets of three numbers a<b<c and d≤e≤f such that a+b+c = d+e+f and abc = def. a, b, and c are the answer, while d, e, and f are there so that Bob's knowing both the product and the sum will yeild two different answers which can be answered by the red-hair statement by showing that none of the three ages can equal one another.

None of the children can be 0 years old. Bob and Tom have not seen each other for 'a while,' yet Bob knows Tom has 3 children. The youngest must have been born before they last met. Had Bob and Tom merely been referring to a month when they said 'while' it is not likely that Bob would have forgotten Tom's newborn baby. Also, pointing to a building with no windows is a little obvious. Finally, the controversy cannot be solved only with the three clues if one of the children is 0, because the product would be 0, resulting in floor([bob's age / 2] - .5) possible answers after all the clues have been disclosed. Unless Bob is 3 or 4 years old, none of the children can be age 0.

Bob is at least 18. Tom is at least 18. Federal rules require at least a 2 year driving record in order to apply for a position as a city mail carrier. Also, both are less than 62 years old, since federal employee retirement kicks in at 62. If we use this info to restrict Bob's age, we set a limit on possible answers. Doing so I get it down to 50 possibilities.

If we say also that Tom was at least 18 when he had his first child and was no more than 50 when he had his last, we can eliminate 5 other possibilities (e.g. Bob is 56, the children are 13, 18, & 25, making Tom at least 63 at the time of the conversation, which is past retirement).

Here is one possibility:

1) Bob is 34. There are 96 different combinations of 3 numbers which sum to 34. Bob doesn't know the answer yet.

2) There is some number of windows on the building across the street which at least two of the 96 combinations share as a product. There are 5 such pairs:

A   B	C	 Product

1   12   21	252

3   3	28	252

1   15   18	270

2   5	27	270

2   14   18	504

3   7	24	504

4   15   15	900

5   9	20	900

6   12   16	1152

8   8	18	1152[/code]
[/font]

There must be at least two with the same product because otherwise Bob would know the answer after the second clue, but he doesn't.  But he does narrow it down to only two possibilities, because he knows how many windows there are, and so he knows the product, and thus which pair to choose.

3) The third clue eliminates the possibility of twins or triplets.  In this case, Bob could only solve the problem if there were 252 or 1152 windows on the building, because this clue wouldn't distinguish between the other three pairs.  So if Bob is 34, the children are either 1, 12, and 21 or 6, 12, and 16.  Of course, since Bob knows the number of windows, he knows which is the right answer.



(In some cases, two sets of ages shared the same product, but both sets also included twins.  Since Bob would not have been able to use the third clue to distinguish them, these can be thrown out, too.)



There are 45 such possibilites for Bob's ages 18-61.  I won't list them here because they take up a lot of room.  I did notice one oddity, though.  Each possibility is the member of a pair (one with twins, one without) with the same product, which product did not recur a third time for that value of Bob's age.  Except one.  At Bob's age 49, there were three sets of ages with the same product and sum:[font=Courier New]


[code]A	B	C	Product

9	20   20   3600

10   15   24   3600

12   12   25   3600

This is the only time in the range of 18 ≤ Bob's age < 62 that a triplet occurs wherein two of the three can be eliminated by the third clue (leaving 10, 15, and 24 as the answer), which makes me think the thing I'm missing has to do with this right here - can we throw out all the pairs because of something I missed? Should I be looking only for triples like this one? What am I missing?

-Q

[jwoolcutt]

Should I be looking only for triples like this one? What am I missing?

I think you're definitely on track. I've been at work today, and don't have Mathematica handy or I would double check your work.

[spoiler warning]

Call a number "ambiguous" if it factors into three numbers in two different ways such that those factors add up to the same number. Make a list of the "ambiguous" numbers up to some reasonable limit (like 200) there is only one "ambiguous" number such that knowledge that there are no "twins" in the factorization determines the factorization That number with that factorization is the one you're looking for and the factors are the ages of the children.

[Qwertz]

So you are saying that there are fewer than 200 windows on the building?

Tom looks around, "Well Bob, the product of their ages is the number of windows on the building across the street."

...

Tom thinks for a minute, "You're right! The sum of their ages is your age Bob."

-Q

[jwoolcutt]

So you are saying that there are fewer than 200 windows on the building?

-Q

Perhaps the word sum and product show up reversed in the text (it's been 3 years since I read the problem). However, I don't think it is significant in solving the problem, since you don't know bounds on the numbers to begin with--the number of windows on the building and the age of the guy are the very last thing you find out, technically, if you solved the problem right. The key is the fact that the postman can't solve the problem until the last statement which determines the solution. There is still only one reasonable triple of numbers that solves the problem.

I will work out the problem tonight on Mathematica and check your solution.

Best,

Jacob

[Capitalism Forever]

WARNING: SPOILER

Make a list of the "ambiguous" numbers up to some reasonable limit (like 200)

Ha! That changes the situation quite a bit! There are only three ambiguous numbers between 1 and 200: 40, 90, and 144. Assuming that the product is indeed Bob's age, rather than the number of windows, and dismissing the possibility of 90+ year-old postmen, we've finally got the unique solution.

(I don't know if there is an elegant way to find the ambiguous numbers, though ... I just asked my computer! )

[Qwertz]

Of course it changes everything. A 14-year-old postman is outside the range of ages I worked with. With the correction, I get the same answer CF does.

-Q

[Capitalism Forever]

BTW, just to quibble a bit, it is quite possible to have a baby and then have another baby 9 months later, so the remark about the middle child does not necessarily imply that the ages of the children are different as expressed in whole numbers of years.

Nonetheless, interesting puzzle!

[blackdiamond]

Would the question "How do you determine the counterfeit and whether it is heavier or lighter in three weightings?" work just as well (i.e., is this a trick question)?

it's not a trick question, Aleph. Yes, you can paraphrase it any way you like. Unbelievable, but it does work!

For practice, try working out one in which you have 27 TENNIS BALLS, ALL HAVE IDENTICAL WEIGHT EXCEPT ONE WHICH IS SLIGHTLY HEAVIER THAN THE REST. HOW DO YOU IDENTIFY THIS HEAVIER BALL IN ONLY THREE WEIGHINGS? Same kind of question except this time there are MORE items (27), but you are told that it is actually heavy (whereas in the other case you are not told, but you must still be able to identify it). If you fail to answer this one (the 27 balls), then you might want to forget about the other one (the 12 coins).

Thanks.

[blackdiamond]

BREAKING NEWS::: CAPITALISM FOREVER HAS SOLVED THE 12 COINS PUZZLE!

[Qwertz]

BREAKING NEWS::: CAPITALISM FOREVER HAS SOLVED THE 12 COINS PUZZLE!

I'd still like to know what was wrong with my solution. I've been over it a dozen times and it still works for me...

-Q

[Qwertz]

PS - Both hunterrose and Mae have submitted correct answers to my zebra puzzle. Congrats to both of them!

-Q