Riddles and Puzzles

[Ducttapers4JC]

There is one solution to this problem: cut a domino in half. (I hope my brother never notices that I worked this problem with his dominos then used super glue to fix it.....just kidding, I never actually cut it in half)

[YoursFalsey]

There is one solution to this problem: cut a domino in half. (I hope my brother never notices that I worked this problem with his dominos then used super glue to fix it.....just kidding, I never actually cut it in half)

As noone has stepped up to explain why the problem is impossible, I choose to give ducttapers4JC credit for an unorthodox solution and invite him to propose the next problem or riddle.

(And for any who had not seen this chestnut before, note this. For any intact domino, when you lay it on the chessboard, it will fall on one red square and one black square. However, both squares removed were the same color (We'll say red, although depending on how you put the chessboard down they could be both black....). Thus after you place the first thirty dominoes, you will have two black squares left- therefore non-contiguous. One domino cannot cover them- unless the domino is cut in half....)

[Ducttapers4JC]

Ok, here is my riddle. It is complicated, but please try to post your work. A publishing company has hired six translators to translate their books into Portuguese, Russian, Swahili, Chinese, French, and Arabic. By coincidence, the last name of each translator begins with a letter that starts the name of one of the languages. Each translator can efficently translate two of the six languages, but no two of them can translate the same two languages. Each language can be translated be exactly two translators. None of the translators can write the language that starts with the same letter as his last name. Mr. S can write Russian and Portuguese. One of the translators can write both Russian and Chinese. Mr. F and Mr. R can write the four languages that do not start with the same letters as there last names. The translators whose last names begin with the same letter as the languages that Mr. R can translate can both write French. Nether of the translators who write Portuguese can write Chinese. Now here is the question. What two languages can Mr. A write?

[YoursFalsey]

Ok, here is my riddle. It is complicated, but please try to post your work. A publishing company has hired six translators to translate their books into Portuguese, Russian, Swahili, Chinese, French, and Arabic. By coincidence, the last name of each translator begins with a letter that starts the name of one of the languages. Each translator can efficently translate two of the six languages, but no two of them can translate the same two languages. Each language can be translated be exactly two translators. None of the translators can write the language that starts with the same letter as his last name. Mr. S can write Russian and Portuguese. One of the translators can write both Russian and Chinese. Mr. F and Mr. R can write the four languages that do not start with the same letters as there last names. The translators whose last names begin with the same letter as the languages that Mr. R can translate can both write French. Nether of the translators who write Portuguese can write Chinese. Now here is the question. What two languages can Mr. A write?

Wow. A doozy. OK- we know one translator speaks both Russian and Chinese, and that isn't Mr F or Mr R (given), Mr C. (given not to speak chinese), or Mr. S (who speaks Portugese and Russian) so that is either Mr. A or Mr. P. Try cases:

Case I: Mr. A Speaks Russian and Chinese. This leads to a contradiction however- Neither Mr A nor Mr S speak French, and likewise, Neither Mr. F nor Mr. R do as they speak the other four languages between them. Thus the french speakers are Mr. C and Mr P, which means Mr. R speaks Portugese and Chinese which no translator speaks both. So Case II must hold-

Case II: Mr. P speaks Russian and Chinese. Mr. S speaks Russian and Port. The two French speakers are Mr. A and Mr. C, so Mr. R speaks Arabic and Chinese, leaving Mr. F to speak Portugese and Swahili. We need one more Arabic speaker and one more Swahili speaker- Mr. A doesn't speak Arabic, so he must be the 2nd swahili speaker leaving Mr C to speak French and Arabic.
So Mr. A speaks French and Swahili. Deep sigh of relief!

[Ducttapers4JC]

Ok, here is my riddle. It is complicated, but please try to post your work. A publishing company has hired six translators to translate their books into Portuguese, Russian, Swahili, Chinese, French, and Arabic. By coincidence, the last name of each translator begins with a letter that starts the name of one of the languages. Each translator can efficently translate two of the six languages, but no two of them can translate the same two languages. Each language can be translated be exactly two translators. None of the translators can write the language that starts with the same letter as his last name. Mr. S can write Russian and Portuguese. One of the translators can write both Russian and Chinese. Mr. F and Mr. R can write the four languages that do not start with the same letters as there last names. The translators whose last names begin with the same letter as the languages that Mr. R can translate can both write French. Nether of the translators who write Portuguese can write Chinese. Now here is the question. What two languages can Mr. A write?

Wow. A doozy. OK- we know one translator speaks both Russian and Chinese, and that isn't Mr F or Mr R (given), Mr C. (given not to speak chinese), or Mr. S (who speaks Portugese and Russian) so that is either Mr. A or Mr. P. Try cases:

Case I: Mr. A Speaks Russian and Chinese. This leads to a contradiction however- Neither Mr A nor Mr S speak French, and likewise, Neither Mr. F nor Mr. R do as they speak the other four languages between them. Thus the french speakers are Mr. C and Mr P, which means Mr. R speaks Portugese and Chinese which no translator speaks both. So Case II must hold-

Case II: Mr. P speaks Russian and Chinese. Mr. S speaks Russian and Port. The two French speakers are Mr. A and Mr. C, so Mr. R speaks Arabic and Chinese, leaving Mr. F to speak Portugese and Swahili. We need one more Arabic speaker and one more Swahili speaker- Mr. A doesn't speak Arabic, so he must be the 2nd swahili speaker leaving Mr C to speak French and Arabic.
So Mr. A speaks French and Swahili. Deep sigh of relief!

Wow YoursFalsey, not only did you work that out fast, but your explination was more clear then mine would have been! Sence you got this right, the next riddle is your to post.

[YoursFalsey]

I've got nothing off the cuff so I am stealing a puzzle:

How many distinct ways are there to colour the six faces of a cube using six different colours, and applying a different colour to each face. Cubes which could be rotated to look the same as each other are not counted as distinct.

(I stole this from http://www.mathmos.net/puzzles/combination.html but there are no solutions posted, so you'll to get one the old-fashioned way- by solving the puzzle!)

[natty dread]

I'm going to guess 120?

[YoursFalsey]

I'm going to guess 120?

As always, I want to see work. (I am a math teacher, so I have an excuse.) Still, guessing at how you guessed, I'm going to suggest that you haven't figured all the ways you can duplicate one cube...

[slowreactor]

I've got nothing off the cuff so I am stealing a puzzle:

How many distinct ways are there to colour the six faces of a cube using six different colours, and applying a different colour to each face. Cubes which could be rotated to look the same as each other are not counted as distinct.

(I stole this from http://www.mathmos.net/puzzles/combination.html but there are no solutions posted, so you'll to get one the old-fashioned way- by solving the puzzle!)

720, or 6!

The 1st face has 6 different options to be painted, the 2nd face has 5, so on, so on.

[jonesthecurl]

I don't think so. let me think about it. But once you colour the first face in colour a, it matters whether you colour an adjoining face in colour b or whether you use colour b for the opposite face.

[jonesthecurl]

I think. Now I'm gonna hand that prob over to the clever bit of me and go to sleep.

[natty dread]

6*5*4*3*2*1 = 720

/6 because there's 6 repetitions of each pattern if you turn the cube around

= 120

no?

[jonesthecurl]

no.
Put colour "a" on side 1 (that's any side at all).
At this point the only colour that matters is the colour on the opposite face - any adjacent face being coloured at this point will be covered later in the prob.
There are five colours which can go on the opposit face. Add colour "b".
The third colour to be added can be any colour - it will be adjacent to "a" and "b" no matter what you do - all solutions are equivalent at this point. Add colour "c" to one of the faces.
So there are 5 ways of adding the first three colours.
Once again, the opposite face is the inportant one - there are three colours which can be added to the face opposite colour "c".
Add colour "d".
Now the addition of the remaining two colours will give the same pattern (after rotation) no matter which way you do it. Add colours "e" and "f".

So if I'm right, and I think I am, there are 5x3 combinations. That's, uh, 15.

[ender516]

no.
Put colour "a" on side 1 (that's any side at all).
At this point the only colour that matters is the colour on the opposite face - any adjacent face being coloured at this point will be covered later in the prob.
There are five colours which can go on the opposit face. Add colour "b".
The third colour to be added can be any colour - it will be adjacent to "a" and "b" no matter what you do - all solutions are equivalent at this point. Add colour "c" to one of the faces.
So there are 5 ways of adding the first three colours.
Once again, the opposite face is the inportant one - there are three colours which can be added to the face opposite colour "c".
Add colour "d".
Now the addition of the remaining two colours will give the same pattern (after rotation) no matter which way you do it. Add colours "e" and "f".

So if I'm right, and I think I am, there are 5x3 combinations. That's, uh, 15.

I think you have reduced it too far. You are right as far as the first three colours are concerned. Imagine "a" on top, "b" on the bottom, and "c" on the front. Now you have three colours to go on the right, back and left sides. Those can be arranged six ways for a total of 5x6 or 30 patterns. You went astray assuming that those last two colours "will give the same pattern (after rotation)". With a, b, c, and d as described, if e is on the right, there is no way to rotate it and put it on the left without swapping some other pair.

[jonesthecurl]

You're right I think. Then there's twice as many possibilities, so, yes, that's 30.

[YoursFalsey]

Ender and Jones are right- 30 possibilities.

For another way of looking at it... Like Jones suggested, put color a on anyside and rotate the cube so that the colored side is the bottom. For the remaining 5 sides, there are 5! ways to arrange the remaining 5 colors. (Color b has 5 possible locations, color c has 4 possible locations, etc. ) That is the 120 natty_dread came up with.

However, even with color a on the bottom, you can rotate a given painted cube 4 ways. (Pick a given side other then top or bottom and rotate it to face North, East, South, or West.) So each possible arrangement is repeated 4 times in the list of 120 arrangements we just made. 120/4=30 so there are 30 arrangements once we compensate for that repetition.

Ender said 30 first, so it is his turn to puzzle us!

[natty dread]

However, even with color a on the bottom, you can rotate a given painted cube 4 ways. (Pick a given side other then top or bottom and rotate it to face North, East, South, or West.) So each possible arrangement is repeated 4 times in the list of 120 arrangements we just made. 120/4=30 so there are 30 arrangements once we compensate for that repetition

ahh, I didn't think it far enough. gg guys

[ender516]

Okay, here's a blast from the past. This puzzle was submitted by Lars Bertil Owe of Lund, Sweden, to Martin Gardner's Mathematical Games Column and I found it republished in the collection Knotted Doughnuts and Other Mathematical Entertainments.

My wife and I recently attended a party at which there were four other married couples. Various handshakes took place. No one shook hands with himself (or herself) or with his (or her) spouse, and no one shook hands with the same person more than once.

After all the handshakes were over, I asked each person, including my wife, how many hands he (or she) had shaken. To my surprise each gave a different answer. How many hands did my wife shake?

[nippersean]

See below for my second attempt

[nippersean]

She shook 4 peoples hands I think - can't do the maths though.

I did it that each couple shook 8 hands between them, but found they shook both mine + my wive's or neither in my e.g. (I just did a grid and tried it). C'mon maths bods where's the proof??

[YoursFalsey]

She shook 4 peoples hands I think - can't do the maths though.

I also like the four answer, but don't have the math to back it up yet. However, consider. There were 10 people at the party, so he asked 9 people how many hands they shook. Possible answers ranged from 0 to 8 (since nobody shook their own hand or their spouse's hand.) Since each of the nine people answered a different answer, it follows each answer was given once.

We also know that if you added the speakers handshakes in, you would get an even number, as every handshake was counted twice, once by each end of the hand shake. Since the sum of 0 through 8 is 36, also an even number, it follows that the husband shook an even number of hands (0, 2, 4, 6, or .

If we could prove each couple shook 8 hands total, the lack of duplicate answers would mean the wife shook 4 hands and the speaking husband shook 4 hands. I don't see how to prove that though.

Two approaches present themselves. One) Assume a couple shook more or fewer hands and see if that leads to a contradiction. Two) See if the givens mean that for each couple, every partygoer shook the wife's hand or the husband's hand but not both. (Which gives us the each couple shook 8 hands total, and we are then there.) I'm going to think about this some more while I watch football- good luck on getting there...

[YoursFalsey]

Now I have the math- figured it out during homeroom this morning!

Somebody told the speaker they shook 8 hands- every hand they could legally shake since spousal-shaking and self-shaking are off limits. Was it the speaker's wife? No, because in that case, each of the other people he asked shook at least one hand- the speaker's wife- and again somebody had to shake 0 hands. So someone (Call that individual A) shook 8 hands. A's spouse is the only person who didn't shake A's hand, so the spouse (Call the spouse a) must be the person who shook 0 hands. So we know one of the couples, A and a, accounted for the 8 and 0 answers.

Somebody told the speaker they shook 7 hands. Was it his wife? If it was, then again we have a contradiction- everyone remaining had to shake hands both with A and with the speaker's wife, yet somebody remains who only shook 1 hand. So someone other then the speakers wife (Call them B) is the person who shook 7 hands. Likewise, B's spouse (b) must be the person who shook hands with A but nobody else. So one of the couples, B and b, accounted for the 7 and 1 answers.

Similar logic will show that C and c answered 6 and 2, D and d answered 5 and 3. The only unaccounted for answer is 4, so that must be the answer the speaker recieved from his wife. (And as it turns out, 4 is also the number of hands the speaker shook.)

Congratulations ender- that was a damn fine puzzle! Thanks also to nippersean, for providing the impetus that started me down the road to the answer.

[ender516]

Now I have the math- figured it out during homeroom this morning!

Somebody told the speaker they shook 8 hands- every hand they could legally shake since spousal-shaking and self-shaking are off limits. Was it the speaker's wife? No, because in that case, each of the other people he asked shook at least one hand- the speaker's wife- and again somebody had to shake 0 hands. So someone (Call that individual A) shook 8 hands. A's spouse is the only person who didn't shake A's hand, so the spouse (Call the spouse a) must be the person who shook 0 hands. So we know one of the couples, A and a, accounted for the 8 and 0 answers.

Somebody told the speaker they shook 7 hands. Was it his wife? If it was, then again we have a contradiction- everyone remaining had to shake hands both with A and with the speaker's wife, yet somebody remains who only shook 1 hand. So someone other then the speakers wife (Call them B) is the person who shook 7 hands. Likewise, B's spouse (b) must be the person who shook hands with A but nobody else. So one of the couples, B and b, accounted for the 7 and 1 answers.

Similar logic will show that C and c answered 6 and 2, D and d answered 5 and 3. The only unaccounted for answer is 4, so that must be the answer the speaker recieved from his wife. (And as it turns out, 4 is also the number of hands the speaker shook.)

Congratulations ender- that was a damn fine puzzle! Thanks also to nippersean, for providing the impetus that started me down the road to the answer.

Yes, you have it just right. That is the very reasoning used in the solutions in the book. Thanks again, Messrs. Gardner and Owe. You're up for the next puzzle, YoursFalsey.

[nippersean]

No problem Yours, just a small point, I didn't provide the impetus to your answer, I provided the answer.

Hurry up then with with the next one!

[YoursFalsey]

No problem Yours, just a small point, I didn't provide the impetus to your answer, I provided the answer.

Details, details...

Hurry up then with with the next one!

OK, a quick and simple one: {stolen from Raymond Smullyan}

The Island of Questioners
Somewhere in the vast reaches of the ocean, there is a very strange island known as the
Island of Questioners. It derives its name from the fact that its inabitants never make
statements, they only ask questions. The inhabitants ask only questions answerable by Yes or
No. Each inhabitant is one of two types, A and B. Those of type A ask only questions whose
correct answer is Yes; those of type B ask only questions whose correct answer is No. For
example, an inhabitant of type A could ask, "Does two plus two equal four?" But he could not
ask whether two plus two equals five. An inhabitant of type B could not ask whether two plus
two equals four, but he could ask whether two plus two equals five.
I once visited this island and met a couple named Ethan and Violet Russell. I heard Ethan
ask some, one, "Are Violet and I both of type B?' What type is Violet?