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July 16, 2005

A small puzzle
Posted by Teresa at 11:00 AM * 129 comments

I ran across a nice puzzle, posted by one Bernard Guerrero in a comment thread on Halfway Down the Danube:

Puzzle: Person x and y have the following conversation:

x: I forgot how old your three kids are. y: The product of their ages is 36.
x: I still don’t know their ages.
y: The sum of their ages is the same as your house number.
x: I still don’t know their ages.
y: The oldest one has red hair.
x: Now I know their ages!

What are the kids’ ages?

Shush, JVP. We know you can solve this one.
Comments on A small puzzle:
#1 ::: Julie L. ::: (view all by) ::: July 16, 2005, 11:56 AM:

Actually, I think there's more than one solution. The last clue seems intended to exclude the oldest child(ren) from being 6-year-old twins, which would've made the third child a one-year-old. The most obvious set is then three-year-old twins and a four-year-old redhead, but it doesn't completely exclude possibilities like an 18-year-old redhead with much younger sibs (1 and 2 years old), unless I'm missing something.

#2 ::: Molly ::: (view all by) ::: July 16, 2005, 11:57 AM:

The kids are 9, 4, and 1

#3 ::: Molly ::: (view all by) ::: July 16, 2005, 11:58 AM:

Julie, what you're missing is that all of the alternate solutions have to add up to the same number (the house number) as the twins at the top solution, ie, 13.

#4 ::: Abigail ::: (view all by) ::: July 16, 2005, 12:04 PM:

Like all numbers, 36 can be broken down into primes in only one way: 36 = 2*2*3*3

From this breakdown we can discover the eight possible configurations of the children's ages:

1,1,36
1,2,18
1,3,12
1,4,9
1,6,6
2,2,9
2,3,6
3,3,4

If we look at the sums of the different configurations, we will find:

1+1+36=38
1+2+18=21
1+3+12=16
1+4+9=14
1+6+6=13
2+2+9=13
2+3+6=11
3+3+4=10

Since we know that person y couldn't tell what the children's ages were from the second hint (and since person y no doubt knows his own house number) we can conclude that only two configurations are possible - 1,6,6 and 2,2,9.

From the third clue, we know that there is an oldest child, so the configuration 1,6,6 is impossible.

Therefore, the children's ages are 2, 2 and 9.

(If you're really impressed, you could, you know, go check out my new blog. If you didn't have anything better to do.)

#5 ::: Greg Ioannou ::: (view all by) ::: July 16, 2005, 12:09 PM:

Neat! The key for me is the "Now I know their ages." That tells me that knowing the house number alone won't give you the answer. The only possible answers where knowing the house number wouldn't tell you which is right are 6-6-1 and 9-2-2, both of which add up to 13. Knowing there is a single eldest child tells you which of those is right.

#6 ::: Greg Ioannou ::: (view all by) ::: July 16, 2005, 12:11 PM:

Abigail, we're both wrong. 1,1,36 is also a right answer, no? OK, I'll do some more thinking.

#7 ::: Greg Ioannou ::: (view all by) ::: July 16, 2005, 12:11 PM:

Never mind, brain cramp. It is 9-2-2.

#8 ::: Abigail ::: (view all by) ::: July 16, 2005, 12:17 PM:

Actually, what I like best about the puzzle is the misdirection of mentioning the kid's red hair. You spend so much energy wondering how hair color could be a factor in a math puzzle that you don't notice the actual information in the clue.

#9 ::: Andy Wilton ::: (view all by) ::: July 16, 2005, 12:19 PM:

At the (fairly high) risk of seeming pedantic, the 6-6-1 solution is not incompatible with there being a single eldest child. Measured in whole years, my wife is the same age as her younger brother for one month every year.

#10 ::: Anarch ::: (view all by) ::: July 16, 2005, 12:24 PM:

There's a similar problem that's one of my absolute favorites....

Two integers are chosen from 2 to 100 inclusive; their sum is given to Sue and their product is given to Patrick. The following conversation ensues:

P: "I don't know what the two numbers are."
S: "Yeah, I knew you didn't know that."
P: "Oh. Now I know what the two numbers are!"
S: "Ah, now I know what the two numbers are too!"

What are the two numbers?

---

Another variation on that theme, the one I originally heard:

P: "I don't know what the two numbers are."
S: "I don't know what the two numbers are."
P: "Now I know what the two numbers are."
S: "Now I know what the two numbers are."

What are the two numbers?

---

The original variation on this problem seems to have first appeared in 1969 but it hit the Internet in a big way in the early '90s on rec.arts.puzzle, I believe. Fun stuff; those who want spoilers (or just to see a collection of its myriad variations) should check this out.

#11 ::: Kylee Peterson ::: (view all by) ::: July 16, 2005, 12:41 PM:

A similar but more annoying problem, which is attributed to G. P. Klimov in my math reasoning book:

Two mail carriers meet on their routes and have a conversation.

A: I know you have three sons. How old are they?
B: If you take their ages, expressed in years, and multiply those numbers, the result will equal your age.
A: But that's not enough to tell me the answer!
B: The sum of these three numbers equals the number of windows in that building.
A: Hmm... But it's still not enough!
B: My middle son is red-haired.
A: Ah, now it's clear!
How old are the sons?

I dislike this one because there seems to be no way of solving other than scattershot examining of some numbers that seem as though they might work. If anyone has a better method, I'd be happy to hear it.

#12 ::: BSD ::: (view all by) ::: July 16, 2005, 12:47 PM:

9 and 2?
P is 18, can be 9x2 or 6x3, P does not know.
S is 11. Possibles are:
9+2 (18 - two possible P pairs)
8+3 (24 - 3 possible)
7+4 (28 - 2 possible)
6+5 (30 - 3 possible)
So S, with 11, knows that no matter WHAT the P is, P does not know, satisfying 2.
Knowing 2, we examine the two possibles of 18: 9x2 and 6x3. If S had the other alternative (9), a possible answer pair would have been 7,2, whose product is 14, a giveaway, so S would not know that P would have a non-revealing number.

I'm probably wrong, I'll check after posting.

#13 ::: Greg Ioannou ::: (view all by) ::: July 16, 2005, 12:52 PM:

Abigail, that was really disconcerting. I went to check out your blog. You've used the same template I've used for mine -- I wondered for a second how I'd been thrown back into my own blog. (I'm having fun exploring yours.)

#14 ::: James D. Macdonald ::: (view all by) ::: July 16, 2005, 01:09 PM:

This reminds me of a puzzle from one of Raymond M. Smullyan's books.

(This is me doing this from memory, not copying from the book, BTW.)

A group of adventurers were gathered at the Explorers' Club talking of their travels.

One of them was speaking:

"As you know, there is a small country in the Balkans where half of the population is vampires. It is well known that all humans always tell the truth, while all vampires always lie. This sorry state of affairs has driven half of the humans insane, so that when they think they're telling the truth they're actually lying. And half of the vampires who were insane humans become insane vampires: When they think they're lying they're actually telling the truth.

"One evening, precisely at twilight when both humans and vampires are abroad, I met a fellow there named Igor. I asked him whether he were a sane human, and he answered 'yes' or 'no,' and from his answer I couldn't tell what he was."

"How odd!" says a second adventurer. "I too traveled in that same country, and precisely at twillight I met that same Igor. I asked him whether he were an insane human, and he answered 'yes' or 'no,' and from his answer I couldn't tell what he was."

"A remarkable coincidence!" says a third adventurer. "I visited that land as well, and precisely at twillight one evening I met that very Igor. I asked him whether he were a sane vampire, and he answered 'yes' or 'no,' and from his answer I couldn't determine what he was."

From the overstuffed chair by the fire, Professor Challenger spoke: "My dear boys! It's perfectly obvious what he was."

#15 ::: Abigail ::: (view all by) ::: July 16, 2005, 01:30 PM:

In each of the three cases, an answer of 'yes' is only possible from one of the four groups.

When asked if they are a sane human, only sane humans will say 'yes'.

When asked if they are an insane human, only sane vampires will say 'yes'.

When asked if they are a sane vampires, only insane humans will say 'yes'.

Since all three travelers couldn't determine which group Igor belonged to, it follows that he answered all three questions with 'no'.

Only one subgroup of the country's citizens will answer all three questions with 'no' - Igor is an insane vampire.

Poor Igor.

(I'm having fun exploring yours.)

Oh, damn, I better go write more entries.

I had a similar experience with the template issue. It's a bit like two women showing up at a party wearing the same dress, except that there are several gajillion guests.

#16 ::: tavella ::: (view all by) ::: July 16, 2005, 01:58 PM:

But even if there are twins, one was still born first, and so there's still an older child. At least, the twins I've known were quite aware of who was the older and who was the younger.

#17 ::: Francis ::: (view all by) ::: July 16, 2005, 02:11 PM:

When asked if they are a sane human, only sane humans will say 'yes'.

Um. Won't a sane vampire realise that he isn't a sane human, and the answer is therefore no - so he will lie and claim he is a sane human?

#18 ::: Teresa Nielsen Hayden ::: (view all by) ::: July 16, 2005, 02:18 PM:

Abigail, the red-haired herring charmed me too. I also liked the elegance of reusing "I still don't know their ages" to exclude all the unique possible answers.

#19 ::: Abigail ::: (view all by) ::: July 16, 2005, 02:31 PM:

Of course you're right, Francis. It should read, When asked if they area a sane human, only insane vampires will say 'no'.

Or, to clarify:

The first question is 'Are you a sane human?'
A sane human is inclined to say 'yes' and because he is sane he will say 'yes'.
An insane human is inclined to say 'no' and because he is insane he will say 'yes'.
A sane vampire is inclined to say 'yes' and because he is sane he will say 'yes'.
An insane vampire is inclined to say 'yes' and because he is insane he will say 'no'.
Since the traveller couldn't determine which group Igor belonged to, we can conclude he said 'yes' and is therefore either a sane or insane human, or a sane vampire.

The second question is 'Are you an insane human?'
A sane human is inclined to say 'no' and because he is sane he will say 'no'.
An insane human is inclined to say 'yes' and because he is insane he will say 'no'.
A sane vampire is inclined to say 'yes' and because he is sane he will say 'yes'.
An insane vampire is inclined to say 'yes' and because he is insane he will say 'no'.
Again, we can conclude that the answer was 'no' and therefore Igor is either a sane or insane human or an insane vampire. From the first clue we now know that Igor must be human.

The third question is 'Are you a sane vampire?'
A sane human is inclined to say 'no' and because he is sane he will say 'no'.
An insane human is inclined to say 'no' and because he is insane he will say 'yes'.
A sane vampire is inclined to say 'no' and because he is sane he will say 'no'.
An insane vampire is inclined to say 'yes' and because he is insane he will say no.
Therefore, we can conclude that Igor must have answered the third question with a 'no' and the only kind of human who does that is a sane human.

Lucky Igor.

#20 ::: Anarch ::: (view all by) ::: July 16, 2005, 02:33 PM:

An insane human is inclined to say 'no' and because he is insane he will say 'yes'.

Ah, but is the insane human aware of his insanity? ;)

#21 ::: Anarch ::: (view all by) ::: July 16, 2005, 02:49 PM:

Oops. Slight correction to the second of the two problems I posted above: in that version you need the sum of the two numbers to be less than 100 -- not just that the numbers are individually less than 100 -- or the answer's not unique.

And Kylee, there is a general sort of strategy involved in these problems, although there's also a hefty amount of brute force as well in every solution I've seen. IOW, a certain cleverness will allow you to say "I'm looking for a number with properties X, Y, Z", at which point you list them all out and see which ones meet the criteria. More of the same cleverness can reduce the amount of work this latter step takes, but it's ultimately gonna be brute force.

Now the really interesting question is: if you lift the restriction on the size of the numbers, is it possible to completely (and usefully) characterize all possible solutions to one of these problems? That would most definitely require cleverness orders of magnitude beyond that required to solve the problems as stated, and damned if I know how to do it.

#22 ::: Jonathan Vos Post ::: (view all by) ::: July 16, 2005, 02:54 PM:

Consider me shushed on the topic as such. After all, I have my copy of Harry Potter 6, and was utterly delighted with the opening chapter. And, you know, San Diego Comic-Con.

Also, look over at Michael Brub's blog: Arbitrary. Fun.
Charlie Harris, professor emeritus of English at Illinois State University, contemporary literature reader/critic extraordinaire (secretary of the Center for Book Culture.org and former director of the Unit for Contemporary Literature), and all-around fine fellow, informs me that a bunch of literary-minded folk are putting together a list of Great First Lines in Novels, as an arbitrary-but-fun counterpart to the American Film Institutes 100 great movie lines.

So far they have over 150 nominations...

As of now, as I post to Making Light, Michael Brub's readers have posted 126 comments, some with many more first lines, and side-comments on some Science Fiction and Fantasy classics.

#23 ::: DBratman ::: (view all by) ::: July 16, 2005, 02:57 PM:

y: The sum of their ages is the same as your house number.
x: I still dont know their ages.

Even if this left a unique answer, if X is not a math whiz X might still not know their ages, at least not off the top of the head. (Although X's response to the third question suggests that X is indeed a math whiz.)

y: The oldest one has red hair.

Does this require that the other two don't have red hair?

#24 ::: James D. Macdonald ::: (view all by) ::: July 16, 2005, 03:08 PM:

My twins are very keenly aware of which one is older. But that isn't the point of the puzzle.

Meanwhile ...

Here's a fun puzzle:

Crossing the Bridge

The Royal Family is in trouble and they need your help!

The sun has set as the King, the Queen, the Knight and the Lady arrive at the bridge right outside the castle walls. They have just returned from a long and dangerous trip across the realm. Scouts have spotted the King's arch enemy closing in on the castle. It is estimated that the King's foe will arrive accompanied by merciless troops in 17 minutes. That means that all four member of our Royal party must cross the bridge and rest safely together on the bridge's other side before disaster strikes!

It's not as easy as it seems. Only two people can cross the bridge at once (it's not as sturdy as it used to be) and those crossing the bridge must carry the torch (so that they can see!). The torch can't be thrown across the bridge... it's windy tonight and the torch would probably blow out... leaving our Royal friends to stumble in the dark.

To make matters worse each person takes a different amount of time to cross the bridge. If two people cross the bridge together it will take them the longer of the two travel times to cross. In other words, the Knight can cross the bridge in one minute, but the Queen requires ten minutes. Therefore, if the Knight and Queen cross the bridge together a precious ten minutes will have past.

Remember that the clock is ticking... there is only 17 minutes for us to figure this little problem out... GOOD LUCK!

The Knight takes one minute, the Lady takes two minutes, the King takes five minutes, and the Queen takes ten minutes to cross that bridge. (It takes those folks the same amount of time to cross back with the torch, too....)

#25 ::: Larry Brennan ::: (view all by) ::: July 16, 2005, 03:21 PM:

Jim - That's a classic problem, and I think you've found a really novel wording.

To those who haven't seen it before - try some non-intuitive combinations.

This problem is often found in references for people who are applying for jobs in the consulting or software industries.

#26 ::: Clifton Royston ::: (view all by) ::: July 16, 2005, 03:53 PM:

Jim: Got it. That one took me a couple minutes; I don't think I've ever seen that particular puzzle before. The key is realizing that the order in which a certain pair of royals goes across is critical. Once you have that, there are two equivalent solutions.

We have two very rambunctious new catlings here (a little too big to be kittens...) When the catlings are in the house, getting dinner from the kitchen to the table turns into a "bridge" puzzle - leave a plate with food unattended on either the counter or the dining table for more than a moment, and there will be a small white face poking into it. When you add the 3-year-old-needs-watching-and-help constraint, and when only the first couple of us have trickled in for dinner, it can require some rapid problem solving to have both locations covered. Hmmm, maybe this would make a fun basis for a puzzle!

#27 ::: Anarch ::: (view all by) ::: July 16, 2005, 04:25 PM:

Another one my friend told me just recently, also in the context of job interviews: you've got a list of 1001 numbers, the numbers 1-1000 with one repetition, that aren't arranged in any particular order. What's the most efficient way to figure out what the repeated number is?

Another one he told me: You again have a list of numbers of some fixed length (say 1000) where the numbers are a) all between 0 and 10, and b) arranged in non-decreasing order (so each number is greater than or equal to the one preceding it). What's the most efficient way to add them up?

#28 ::: Clifton Royston ::: (view all by) ::: July 16, 2005, 05:13 PM:

Jim -

I have a devilish twist for you - I thought this one up in the shower just now.

The royal party has quickly figured out a solution - all that chess at court has sharpened their problem-solving skills - but just as they are about to cross, the Queen's young Page runs up. He tells them that the bridge has been strengthened recently, but only to the point that it can support 3 women and youths at once - having a full-grown man on it with two other people would be too dangerous.

Now they must get the Page across too. The Page can cross just as quickly as the Knight. The enemy army is still 17 minutes away. Can they still do it?

#29 ::: Clifton Royston ::: (view all by) ::: July 16, 2005, 05:37 PM:

One more, while I'm on a roll:

As with the last variation, but the Page also tells them he twisted his ankle while decoying the enemy arm to delay them a few minutes. With the twisted ankle, it will take him 3 minutes to cross the bridge.

What's the fastest solution to get them all across now?


#30 ::: Carlos of Halfway down the Danube ::: (view all by) ::: July 16, 2005, 05:39 PM:

I think I'll have to put more math up on the blog.

Incidentally, if you're coming for the math, stay for the Balkans! (And Brooklyn.)

#31 ::: Matt McIrvin ::: (view all by) ::: July 16, 2005, 07:36 PM:

I got the answer to the vampire puzzle immediately without working it out in detail. Assuming that Professor Challenger has a correct and unique answer, the answer must be "an insane vampire", by symmetry.

(or is that reasoning correct?)

#32 ::: Matt McIrvin ::: (view all by) ::: July 16, 2005, 07:44 PM:

You know, I like these puzzles, but if I'd had to survive those job interviews where you do logic puzzles under pressure with an interviewer staring at you, I'd be eating out of dumpsters today.

#33 ::: Jonathan Vos Post ::: (view all by) ::: July 16, 2005, 09:42 PM:

Here's one I put on Kathryn Cramer's blog, which fooled her. If you put ALL the prime numbers in alphabetical order, which one comes first?

#34 ::: adamsj ::: (view all by) ::: July 16, 2005, 10:20 PM:

You can't fool me, Jonathan--one isn't a prime number, and so it can't come first.

I'll mention this when I go to my advanced calculus class Monday at, um, eleven?

#35 ::: Jonathan Vos Post ::: (view all by) ::: July 16, 2005, 10:31 PM:

Hint number one:

Sequence A000052 of the more than 100,000 searchable sequences listed at the On-Line Encyclopedia of Integer Sequences is:

8, 5, 4, 9, 1, 7, 6, 3, 2, 0, 18, 80, 88, 85, 84, 89, 81, 87, 86, 83, 82, 11, 15, 50, 58, 55, 54, 59, 51, 57, 56, 53, 52, 40, 48, 45, 44, 49, 41, 47, 46, 43, 42, 14, 19, 90, 98, 95, 94, 99, 91, 97, 96, 93, 92, 17, 70, 78, 75, 74, 79, 71, 77, 76, 73, 72...

Name: 1-digit numbers arranged in alphabetical order, then the 2-digit numbers arranged in alphabetical order, then the 3-digit numbers, etc.

Example: eight, five, four, nine, one, seven, six, three, two, zero, eighteen, etc.

#36 ::: Aquila ::: (view all by) ::: July 17, 2005, 12:55 AM:

Seems to me that it really depends on what alphabetization rules you use. I'm inclined to string together some rules from the Chicago Manual of Style and say the answer would be 101.

But that's partly because I really don't want to know if there are primes that begin eight billion ...

#37 ::: Teresa Nielsen Hayden ::: (view all by) ::: July 17, 2005, 02:10 AM:

What comes next in this sequence?

59, 125, 145, 168, 175, 181, 190 ...

#38 ::: Jonathan Vos Post ::: (view all by) ::: July 17, 2005, 02:14 AM:

Aquila:

You're getting warmer.

Teresa:

It's more fun for me to watch others play, so I'll bite my tongue for a while.

#39 ::: Teresa Nielsen Hayden ::: (view all by) ::: July 17, 2005, 02:17 AM:

Figured I'd get a giggle out of you with that one.

#40 ::: Larry Brennan ::: (view all by) ::: July 17, 2005, 02:31 AM:

TNH - I learned that sequence with slightly different values, but it's a blast from the past nonetheless.

#41 ::: Bob Oldendorf ::: (view all by) ::: July 17, 2005, 02:39 AM:

Ooo! Ooo! I know! I know!

(Partially because that integer sequence has been used as a hypothetical example of how one could build a more culturally-diverse SAT test, and partially because I've used it.)

#42 ::: Peter ::: (view all by) ::: July 17, 2005, 02:40 AM:

Another one my friend told me just recently, also in the context of job interviews: you've got a list of 1001 numbers, the numbers 1-1000 with one repetition, that aren't arranged in any particular order. What's the most efficient way to figure out what the repeated number is?

Sum the numbers and subtract 500,500

#43 ::: Francis ::: (view all by) ::: July 17, 2005, 08:08 AM:

If you put ALL the prime numbers in alphabetical order

After you, and wake me when you've finished.

#44 ::: adamsj ::: (view all by) ::: July 17, 2005, 08:15 AM:

Dammit. Okay, Jonathan, thanks for the hint. As usual, I thought too quickly.

#45 ::: Lisa Spangenberg ::: (view all by) ::: July 17, 2005, 08:34 AM:

Someone who knows Blondie well, wanna give me some album titles to check out? I'm looking for a starting-place . . .

#46 ::: Bill Blum ::: (view all by) ::: July 17, 2005, 11:14 AM:

Despite having dealt with Fourier series, wavelets, and whatnot-- I still grin everytime I hear my favorite integer sequence....

The Pinball Counting Song (from Sesame Street).

#47 ::: Anarch ::: (view all by) ::: July 17, 2005, 11:25 AM:

On which note, I love this little remix, partly because it's so obviously done with love.

#48 ::: Jules ::: (view all by) ::: July 17, 2005, 11:43 AM:

What comes next in this sequence? 59, 125, 145, 168, 175, 181, 190 ...

Nothing: that's the entire sequence "thousands of words in Harry Potter novels".

My favourite puzzle is this one:

On a certain ship, there are 50 pirates, each of which is extremely greedy (that is, given any choice, he will always favour the action that gets him the most money), cutthroat (given a choice, he will always favour the action that involves killing the most people) and perfectly logical. But much more importantly than any of these, they'd rather stay alive for as long as possible. They are also all aware that all of the other 49 have the same attributes. They are ranked in the order in which they joined the crew.

These pirates have just captured a treasure chest, which contains 100 gold pieces. And they have a procedure to determine how to split it: the most senior pirate will propose a way of splitting it, and then all the pirates will vote on it. If half or more of them agree to it, the plan will be used; otherwise the pirate who proposed it will be made to walk the plank, and the next pirate in seniority will propose a plan. This will continue until the loot has been split.

How will the loot be split?

#49 ::: adamsj ::: (view all by) ::: July 17, 2005, 11:45 AM:

Lisa, the first two are must-haves: Blondie and Plastic Letters. Those are great, beginning to end. If there's a nice collection with "Call Me", that'd do, too.

#50 ::: Larry Brennan ::: (view all by) ::: July 17, 2005, 01:09 PM:

Jules - The alternative sequence, from 145 is:
145, 155, 161, 167, 170... It's equally correct.

The way I learned it, though was

4, 14, 23, 34, 42, 50... What's next? And the answer was presented in non-numeric form for maximum cognitive dissonance (and yes, this has two correct answers, too.)

#51 ::: pericat ::: (view all by) ::: July 17, 2005, 01:10 PM:

I rather like Eat to the Beat, also.

#52 ::: pericat ::: (view all by) ::: July 17, 2005, 01:12 PM:

4, 14, 23, 34, 42, 50... What's next?

Hike!

#53 ::: Jon Meltzer ::: (view all by) ::: July 17, 2005, 01:55 PM:

4, 14, 23 ...

I'm stumped. Can I ask my friends Duke and Billy for help?

#54 ::: Jonathan Vos Post ::: (view all by) ::: July 17, 2005, 02:01 PM:

I'm exerting superhuman restraint here, y'know. I've devised and published several hundred math puzzles in the past year and a half. Most would not be right for this blog. My motivation came, in part, from 2 years of teaching college Math to non-abstract non-verbal high-visual students. I dream up a problem more than once day, usually solve it, elaborate on it, tie it to the print Literature and hotlinks, then move on. Sometimes I'm the first to crack a posted math puzzle that baffles greater minds than mine. Sometimes I'm posting helpful comments on someone else's Math. Sometimes the Math puzzles explode into supertar status. Check out, for instance: Sudoku.

A much easier puzzle: explain how the following apparently false statement can actually be true: "My younger brother is older than I am."

As with my previous puzzle, selected for this audience, the issue is more about Enghlish than Math.

Answer to this new one:
Eric W. Weisstein. "Buchowski Paradox."

And, no, he wasn't the heavy-drinking L.A. poet who wrote for sleazy men's magazines and is a huge hit in gritty pseudorealist Europe.

#55 ::: Mary Aileen Buss ::: (view all by) ::: July 17, 2005, 02:25 PM:

A much easier puzzle: explain how the following apparently false statement can actually be true: "My younger brother is older than I am."

It's true of me, in fact, although don't usually phrase it like that. Both of my brothers are older than I, so the younger of the two is still older than I am.

--Mary Aileen

#56 ::: Clifton Royston ::: (view all by) ::: July 17, 2005, 02:46 PM:

Lisa, my first vote would be for a later Blondie album _Autoamerican_, one of my all-time favorite albums. As a bonus, the remastered reissue has the long version of 'Call Me' as one of the extra track. (Seems like everybody has a different favorite, always a good sign.)

#57 ::: Teresa Nielsen Hayden ::: (view all by) ::: July 17, 2005, 03:29 PM:

Sure. And my two older brothers are both younger than I am.

Larry: 59. And if the last integer weren't 50, the answer wouldn't be unique.

#58 ::: Larry Brennan ::: (view all by) ::: July 17, 2005, 03:37 PM:

TNH - the second valid answer is 7.

#59 ::: Jonathan Vos Post ::: (view all by) ::: July 17, 2005, 03:51 PM:

Another one (hint, hunt, more about letters than numbers), but also about primes. This one brought to you by the man who invented "the Game of Life."

Which number comes next, and why?

5, 2, 3, 13, 23, 37, 83, 233, 283, 383, 887, 2383

#61 ::: Lucy Kemnitzer ::: (view all by) ::: July 17, 2005, 04:58 PM:

JVP: are they primes arranged in order of the number of Roman numerals needed to express them?

I don't know what comes next, in any case.

V,II,III, XIII,XXIII, XXXVII,LXXXIII . . . and there I get bored and irritated.

#62 ::: Jonathan Vos Post ::: (view all by) ::: July 17, 2005, 05:12 PM:

Lucy Kemnitzer:

You got it! You actually had it before boredom and irritation set in.

This is my A105269 entry in the On-Line Encyclopedia of Integer Sequences.

"Long" prime Roman numerals. Smallest prime whose Roman numeral representation has n characters. [for n = 1, 2, 3, ...]

The next value is 2887. The sequence is "Finite because of ambiguity of representation for n > 3999. This is the prime version of the sequence A036746 defined by John H. Conway."

5=V, 2=II, 3=III, 13 = XIII, 23 = XXIII, 37 = XXXVIII, 83 = LXXXIII, 233 = CCXXXIII, 283 = CLXXXIII, 383 = CCCLXXXIII, 887 = DCCCLXXXVII, 2383 = MMCCCLXXXIII, 2887 = MMDCCCLXXXVII.

John Horton Conway's sequence is:

A036746:
1, 2, 3, 8, 18, 28, 38, 88, 188, 288, 388, 888, 1888, 2888, 3888
"Numbers with 'long' representations in Roman notation: given by last n letters from ...MMMDCCCLXXXVIII.

See also:

Gerard Schildberger, The numbers from 1 to 3999 expressed as Roman numerals.

#63 ::: Jonathan Vos Post ::: (view all by) ::: July 17, 2005, 05:53 PM:

I should mention this, which everyone who needs, already uses, but some who might enjoy might not know:

Factorization using the Elliptic Curve Method. One of the "alpertrons" on Dario Alpern's site.

Type in a positive integer -- OR an algebraic expression that interprets to an integer -- and this VERY quickly cracks it into its prime factors, as well as its representation as a sum of squares, and some other stuff. It works even for very big numbers.

Since prime numbers are the building blocks of all the whole numbers, this is the great deconstructor, which helps to reveal the Hidden Pattern underlying reality.

#64 ::: Jules ::: (view all by) ::: July 17, 2005, 06:26 PM:

JVP: My younger brother is older than I am.

Then clearly you've been on a trip at a large fraction of c and returned home. ;)

So nobody wants to be a pirate then? Can't say as I blame you.

#65 ::: Lenore Jean Jones ::: (view all by) ::: July 17, 2005, 09:25 PM:

Jules - I was going to guess that the entire prize would go to the least senior pirate. They would all vote to kill off the proposing senior pirate both in order to kill someone and to decrease the possible number of people sharing in the prize. I'm not sure how the fact that all of the pirates would know this factors in, though.

#66 ::: Alan Hamilton ::: (view all by) ::: July 17, 2005, 11:10 PM:

I was thinking along the same way, but that the pot would go to the second least senior pirate as the voting carries with half or more, not a majority. With two pirates left, the second to last can propose all the money goes to himself, and the vote will split 50-50.

I still have no clue on the 4, 14, 23, 34, 42, 50 series. I have to wonder if I posted a series where I randomly picked numbers if anyone would be able to tell, or would I always get a "next number" answer that could be explained one way or another.

#67 ::: Jonathan Vos Post ::: (view all by) ::: July 17, 2005, 11:19 PM:

Alan Hamilton:

Technically, any finite sequence of integers has an infinite number of polynomials which will spit out exactly that sequence in that order, and then go on. That might be fine for a transfinite God who disdains adjectives (another thread), but we humans insist on a meaningful formula which is, in some sense, "simpler" than just the list of integers of the sequence. This quickly leads to rather deep, and in some cases, unsolved, problems in Metamathematics. But this is not the time nor place for me to start explaining Algorithmic Complexity Theory and Gregory Chaitin's astonishing ideas.

#68 ::: Jonathan Vos Post ::: (view all by) ::: July 17, 2005, 11:24 PM:

Whoops. Messed up the start of the YRL. Try:

Gregory Chaitin's astonishing ideas.

Most numbers are random, but it took generations before we could produce a provably random number. Most URLs are random too, and half of those seem to be names of garage bands.

#69 ::: Larry Brennan ::: (view all by) ::: July 17, 2005, 11:42 PM:

Alan Hamilton - The sequence is not made-up, although it's underlying cause does have some randomness. It's familiar to millions of people and actually put to real-world use 24 hours a day, seven days a week.

Here's a hint. This sequence is brought to you by the letters C and E.

#70 ::: Jonathan Vos Post ::: (view all by) ::: July 18, 2005, 12:47 AM:

Re: 4,14,23,34,42,50, ...

The letter A brings you some that start the same and end up with 181, 190, 200, 207.

Now, it so happens that:

181 is prime
190 = 2 x 5 x 19
200 = 2 x 2 x 2 x 5 x 5
207 = 3 x 3 x 23
but this will NOT help you...

#71 ::: Jonathan Vos Post ::: (view all by) ::: July 18, 2005, 12:56 AM:

And it might be even more confusing if I cite the variant associated with #9 that starts the same, then goes from 207 as follows:

207 = 3 x 3 x 23
215 = 5 x 43
225 = 3 x 3 x 5 x 5
231 = 3 x 7 x 11
238 = 2 x 7 x 17
242 = 2 x 11 x 11

#72 ::: Daniel Martin ::: (view all by) ::: July 18, 2005, 01:57 AM:

A further hint would be that the popularity of this sequence (and the assumption that it's fair game to give it) is yet another example of the unconscious cultural imperialism that afflicts the inhabitants of certain famous locations. As is the idea that "A", "C", and "E" could be hints. (If you're familiar with the necessary background info, you probably knew it from when "59" was mentioned next to "125", or at the very least when the sequence "34,42,50" popped up. If you're not, the "A,C, and E are hints" does jack squat.)

Applying my own bit of cultural imperialism, can you tell me which of the following four numbers doesn't belong?

3 7 11 61

Hint: hundreds of people each day see three of these numbers together, without the fourth. Of course, the specific background knowledge that gives us this sequence isn't as hip as that which gives us the previous sequence, so it's not cited several places in lists of "lateral thinking" problems.

As another example, these days I frequently deal with this sequence:

333, 339, 340, 343, 351, ___

What goes in the ___ spot? And what significance does the number 352 have in relation to this sequence?

#73 ::: Larry Brennan ::: (view all by) ::: July 18, 2005, 02:33 AM:

JVP - The letter "A" is only a part-time sponsor of the sequence, and when it is, there's lots of other elements between the 59 and 168.

Daniel Martin - Cultural imperialism? If only! :-)

Sure, there's a group of people who will recognize the sequence instantly, and others who won't. The point is really that not all sequences are rooted in the numerical relationship between their elements, and that we shouldn't limit our search for answers to underlying patterns.

BTW, I'd guess that your sequences are highway exit numbers somewhere, and the numbers in the 300 range would indicate that you're in one of the larger states. Since California only recently started putting exit numbers up, I'd provisionally rule that one out.

#74 ::: Bob Oldendorf ::: (view all by) ::: July 18, 2005, 02:41 AM:

I suppose here is as good a place as any to link to ATT's ever-entertaining Online Encyclopedia of Integer Serquences.

Yes, they even have the answer for Teresa's "59, 125, 145, 168, 175, 181, 190 ..." sequence.

But Daniel Martin's "333, 339, 340, 343, 351" stumped it.

#75 ::: Jonathan Vos Post ::: (view all by) ::: July 18, 2005, 02:54 AM:

Bob Oldendorf:

Since you bring this up (only one letter off, in Hamming Distance), and similarly to looking at the new edits of Wikipedia, there are (myself included) some hardcore types who look every day at:

The On-Line Encyclopedia of Integer Sequences, Recent Additions

"This is a section of the main database for the On-Line Encyclopedia of Integer Sequences. It shows the most recently added sequences in reverse chronological order."

In that sense, it is a kind of multiauthor blog. We like to see each others' work as a preview, of sorts. This way, we make comments that, if acceptable, find their way (with author, email, and date in the attribution) when the commented-on sequence is merged into the main database of 100,000+ sequences. There was a competition to be sequence number 100000. The winner is a real classic...

There is a distinguished editorial panel. Dr. Neil J. A. Sloane is akin to the webmaster. He helpfully comments some sequences "obs" for "obscure" if he doesn't get it, and (very rarely) his highest praise: "nice." He has been known to refer to mine with the even rarer critical term "nonsense." Think of him as the Teresa and Patrick of Mathematics.

#76 ::: Steve Taylor ::: (view all by) ::: July 18, 2005, 03:06 AM:

Lovely puzzle, thank you.

A thousand apologies if the following originally came from Making Light - I've lost track of where I got it:

http://www.drunkmenworkhere.org/170

BE WARNED - idly clicking on answers on this JavaScript driven page will reveal whether you guessed right or not - so don't accidentally reveal the answers to yourself.

First time I did this I got a result which satisfied me, but disagreed with
their answers, and found that their answer to q.20 (which is a pretty dodgy
question) differed from mine - which affected everything else. Got it right
next time, knowing the answer they wanted to q.20.

So - SPOILER - answer to q.20:
q.20 = E

#77 ::: Melanie S. ::: (view all by) ::: July 18, 2005, 05:39 AM:

Although I'm sure you'll all solve it right away, my favorite of the sort of logic puzzle given above has always been:

A king decides to hire the smartest man in the kingdom to he his advisor. After a series of tests, he has the prospects narrowed down to three, so he devises the following puzzle for the men to solve. (I have just realized that this is rather gender-skewed; please feel free to assume these are three women, or three Gethenians, or anything else, as you please.)

Each man is taken into a separate room and has a cross painted on his forehead. The cross may be black or white. Then the king puts the men together in a room. They are to raise their hands if they see two black crosses or a black and a white cross. There are no mirrors in the room or in the hallways leading there; the paint cannot be rubbed off on anything which the men could then see. The men are not allowed to speak to each other. The job, of course, is for each man to figure out which color cross is on his forehead.

Each of the three men have black crosses painted on their foreheads. Of course, when they get into the room, all raise their hands; after a length of time, one man smiles, leaves, and tells the king he has a black cross; he is hired. The question is, how did he figure it out?

(Interestingly, this puzzle is harder knowing the answer already, or so my informal surveys have shown. I think it has to do with the direction from which people approach the problem, eliminating or creating possibilities.)

Also, http://home.cwru.edu/~jnt5/Planarity/ is quite fun for the more spatially oriented...although it will eat your free time if you enjoy it.

#78 ::: Daniel Martin ::: (view all by) ::: July 18, 2005, 06:24 AM:
Sure, there's a group of people who will recognize the sequence instantly, and others who won't. The point is really that not all sequences are rooted in the numerical relationship between their elements, and that we shouldn't limit our search for answers to underlying patterns.
With that basic principle I have no qualms; hence sequences such as this one:
4, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3, ...
Or this
1001, 30, 21, 14, 13, 12, 11, 10, 9 ...
Or this, which was making the rounds a few years back:
71, 42, 12, 83, 54, 24, 95, ...

The difference is that none of those depend on knowledge which would rightly be considered location-specific trivia not appropriate for a general audience were the location not the city that never sleeps. The issue I have with the earlier sequence is that it doesn't actually induce one to stop and try alternate approaches - either the sequence is instantly familiar, because one possesses the background knowledge required, or it is not. I cannot imagine someone who possessed the necessary background knowledge not recognizing "34, 42, 50" or "59, 125". One just doesn't see "59" that often.

However, everyone reading this blog should have the background information necessary for the first and third sequences I just gave, and almost everyone for the second. (And if they don't, it's not because of where they live)

And of course since people here aren't often in the same places I am, the two location-specific groups of numbers I gave in my previous post mean nothing. If you wanted to track them down, you'd have to work out what you can from the information I've posted on this thread, and consider where in the past I might have come across several numbers in the 1-100 range grouped together, and where I'd see numbers in the low- to mid- 300s.

Finding out about me is a silly way to waste your time, and I wouldn't suggest it. Suffice it to say that those sequences depend on information that I believe to be just as obscure as the information behind "4, 14, 23, 34, 42, 50", except that it is tied to less generically hip locations.

#79 ::: candle ::: (view all by) ::: July 18, 2005, 09:58 AM:

Hmm, I'm not very good at maths puzzles and logic puzzles, although I used to be: I think I just no longer have the patience to work through the possibilities, given that an answer clearly exists and is known by the person setting it. Or perhaps that's just a way of saving myself the irritation I would feel at trying and failing. I don't live in NYC either, but then I don't recognise any of Daniel Martin's sequences. But the principle of location-specific knowledge may be admissible if there is a broad enough hint to that effect somewhere in the puzzle. I remember being given this sequence - not to guess at the next term, but to explain what it represented:

HE, HC, HW, HC, HT4, HT123, HC, HW, HC, HE

(I suppose there may be more than one valid answer to this too. It reminds me of a contest to find the most plausible next term in this sequence: G, G, G, G, G, ... - with explanations, please. In this case, I don't have an answer in mind.)


#80 ::: candle ::: (view all by) ::: July 18, 2005, 10:03 AM:

As for James D. MacDonald's royal puzzle, I first came across that - it was trailed as an interview question for Microsoft - in a bizarre form in which the members of U2 had to cross a ravine in order to get to their concert on time. Carrying a torch, obviously. Again, Bono could move at one rate, the Edge at another, Larry more slowly and Adam more slowly still. The supposed solution was the same as you've found for the fairytale form.

But when I first read it, I assumed the answer would be something like:

"All the members of U2 go and look at the ravine. They all stand on the edge. Then the Edge walks across carrying them."

(This answer would be better given aloud, of course, rather than written down.)

Does anyone know if this was in fact an invited answer? If not, I'm not sure what this says about me. I wonder if Microsoft would have hired me.

#81 ::: James D. Macdonald ::: (view all by) ::: July 18, 2005, 11:00 AM:

My own first solution to the Royals Cross the Ravine puzzle was "The Knight and Lady cross the bridge carrying the torch and become the new King and Queen."

--------

It reminds me of a contest to find the most plausible next term in this sequence: G, G, G, G, G, ... - with explanations, please.

O, F, S.

(Go! Go! Go! Go! Go! Oh f**k! Stop!)

---------------

This puzzle was mentioned in another thread here at Making Light:

http://www.drunkmenworkhere.org/170

#82 ::: Jonathan Vos Post ::: (view all by) ::: July 18, 2005, 11:13 AM:

Daniel Martin:

Number of letters in the English name of n, excluding spaces and hyphens, for n = 0, 1, 2, 3, ....

"English names for the numbers from 0 to 1022." by Gerard Schildberger.

The second one is cuter, and I note that
1001 = 7 x 11 x 13.

The third one continues through 20 elements as:

71,42,12,83,54,24,95,66,37,7,78,49,19,81,5,11,21,19,12,61,...

Human beings evolved to find patterns. It is part of what drives the soul of Mathematicians, Poets, Gamblers, and Bloggers.


#83 ::: Daniel Martin ::: (view all by) ::: July 18, 2005, 11:26 AM:
The second one is cuter, and I note that
1001 = 7 x 11 x 13.
Except, of course, that part of the point is that in context it doesn't.
#84 ::: Jonathan Vos Post ::: (view all by) ::: July 18, 2005, 11:49 AM:

Daniel Martin:

Misdirection, on my part. Of course I know that 1001 = 11 x 11.

I am contemplating the sequence:

1001, 100, 21, 14, 13, 12, 11, 10, 9 and wondering why you have a 30 where I have a 21 in such obviously related sequences.

#85 ::: Jonathan Vos Post ::: (view all by) ::: July 18, 2005, 11:50 AM:

I mean "a 30 where I have a 100..."

#86 ::: candle ::: (view all by) ::: July 18, 2005, 01:12 PM:

This puzzle was mentioned in another thread here at Making Light:

Damn, I was hoping not to be repeating stuff. Anyway, I like to think that the next term is 'G', and that it represents a three year old naming the first three horses past the post...

Sorry for any (other) unnecessary duplication.

#87 ::: Daniel Martin ::: (view all by) ::: July 18, 2005, 02:21 PM:

I meant 100, not 30, and was posting with insufficient sleep.

#88 ::: HP ::: (view all by) ::: July 18, 2005, 02:39 PM:

Regarding Theresa's number sequence: Does it make a difference that I've only been to New York City twice in my whole life, and only spent a tiny amount of time navigating the subway on my own? Because if so, I think I've got it figured out, but I don't know the next number in the sequence off the top of my head. I could probably look it up, though.

#89 ::: James D. Macdonald ::: (view all by) ::: July 18, 2005, 02:45 PM:

HE, HC, HW, HC, HT4, HT123, HC, HW, HC, HE

While I don't know what the "H" stands for, I bet that E stands for East, C stands for Center, and W stands for West; T123 is Terminals 1, 2, and 3, while T4 is Terminal 4.

#90 ::: Alex Cohen ::: (view all by) ::: July 18, 2005, 02:48 PM:

While I don't know what the "H" stands for

Given the clusters of 123 and 4, my guess is Heathrow.

#91 ::: Jonathan Vos Post ::: (view all by) ::: July 18, 2005, 02:51 PM:

Also about the amazing On-Line Encyclopedia of Integer Sequences, there are sequences marked "hard" -- meaning that nobody yet knows what the next one in the sequence is, although we all know there must be one.

An interesting presentation is through WebCam for On-Line Encyclopedia of Integer Sequences
Point camera at:
[ ] Recent additions
[ ] Best sequences
[ ] All sequences
[ ] Sequences that need extending

Refresh every [ ]5 [ ]8 [ ]12 seconds, or [ ] when I say so

Click here to start camera [ ]

Also, since I mentioned it, it's fun to see photos of the frequent contributors, some rather famous Mathematicians, some obsessed geeks, hard to tell apart sometimes without a program:
This is a record of the e-party (August-December 2004) to celebrate the arrival of the 100,000-th sequence A100000

(The party is now over - but this record will remain.)

The OEIS 100K E-Party (Page 1)
Welcome to the e-party to celebrate the arrival of the 100,000-th sequence A100000!

This is also the 40-th anniversary of the birth of the database (when I was a graduate student at Cornell in the 1960's).

Since thousands of people from all over the world use the OEIS every day, a real party would be impossible. So instead let's have an e-party!

Everyone who uses the OEIS is invited to join!

[I think this gets at the social structure of users orthogonally to Erdos Number, and provides a fascinating glimpse of a subculture alien to most muggles, as it were]


#92 ::: candle ::: (view all by) ::: July 18, 2005, 03:21 PM:

While I don't know what the "H" stands for

Given the clusters of 123 and 4, my guess is Heathrow.

Yep - this is what I meant by giving enough of a clue (not the case with the NYC version). For the rest you will have to look at a map of the London underground. Not every H stands for Heathrow, and not every C for Central. (It is also the longest round trip possible on the underground in which every station begins with the same letter.) Still a bit location-specific, I know.

#93 ::: Mary Kay ::: (view all by) ::: July 18, 2005, 05:54 PM:

There's going to be a Clarion West party at our house Friday night and all of a sudden I'm wondering if it would be evil to serve high octane limeade....

Hmmm.

Say guys, do you know if Gordon Van Gelder likes limeade? She asked innocently.

MKK

#94 ::: Jonathan Vos Post ::: (view all by) ::: July 18, 2005, 06:07 PM:

Here's a recently published puzzle of mine which is similar to Daniel Martin's. What comes next?

10, 101, 1211, 14321, 150145, 266124, 1007614, 1805161, 1805161, 629043, 264043, 93250

On my earlier puzzle, the alphabetically first prime is:

8018018851

Conway and Guy note that 8018018851 is, alphabetically, the first prime number in the American system of large number terminology, and term this "Knuth's number." We have "eight billion" alphabetically before "eighteen" or "eighty."

Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag, p. 15, 1996.

I extended this by asking: what is the alphabetically first n-almost prime for n = 2, 3, 4, 5, 6, 7, 8, 9, 10, ... where, for instance, a "2-almost prime" means a semiprime, namely a number which is the product of exactly two primes (not necessarily distinct). This is a fairly tricky problem, but one part can be done by Making Light readers in their heads: what is the alphabetically first 3-almost prime? That is, of all the infinite (American)-English-language names of whole numbers, which one, that has exactly three prime factors, comes first?

#95 ::: Daniel Martin ::: (view all by) ::: July 18, 2005, 09:16 PM:

Well the 3-almost prime is especially easy since you told us in your earlier post what the alphabetical order was.

And you other sequence is cute too, although I would have cut it off one number earlier, so that one could answer the question "what is the next number" properly. (Though I will say that the 20th term in that sequence is "27", if you look at things the right way)

Oh, by the way: whoever guessed highway exit numbers before is correct - those are the highway exit stops along the PA turnpike across the north end of Philadelphia. The two remaining numbers are 358 and 359, and 352 is the location of the last rest stop. The three numbers which belonged together were 3, 11, and 61. Those three bus routes run up and down Charles and St. Paul Streets in Baltimore. The 7 bus is in a completely different part of the city.

These two were deliberately meant to be as dependant upon obscure location-dependant trivia as the subway stop sequence. There's really no rational reason anyone should have gotten them.

#96 ::: Jonathan Vos Post ::: (view all by) ::: July 18, 2005, 09:27 PM:

Daniel Martin:

Which is why I wrote "What comes next?" instead of "what is the next number?" But you're right, and I see your point. Is there another sequence of that form at least as long, which starts with some other number? If not, it's merely a curiosity that I stumbled on by accident, sent to OEIS, and expected to disappear like a feather dropped into the Grand Canyon, with no echo.

I invented a pseudo-Logic puzzle in the 1960s, which is (I think) actually meaningless. But it has the format and style of a logic puzzle, and I got some funny responses when I asked it at certain parties:

True or false:

"If you don't know
that you know
that you don't know
that you're not stoned,
you're stoned,
but you don't know it."

#97 ::: Teresa Nielsen Hayden ::: (view all by) ::: July 18, 2005, 10:59 PM:

HP, the integer which follows "190" is "Dyckman."

#98 ::: Jonathan Vos Post ::: (view all by) ::: July 18, 2005, 11:47 PM:

Here's another of my number puzzles which is really about the American English name of the number. What number comes next?

2, 0, 11, 14, 102, 101, 111, 114, 1014, 1102, 1101, 1111, 1014, 11114, 14114, 101111, 101114, 111114

The point of most of my puzzles given here is that they are nonthreatening to the math-adverse, as they are actually about language. You know, "Math for Poets." Sometimes the writer/reader can outdo the professional mathematician, who is stuck with thinking "inside the box." That is also part of the value of the subway, freeway, and bus stop puzzles.

#99 ::: Jonathan Vos Post ::: (view all by) ::: July 19, 2005, 12:43 PM:

I was hoping that someone would write out: two, zero, eleven, fourteen, and so forth, and see the pattern in that delightful "aha" experience.

But, for the above, not just a hint, but the entire answer, with a layout that makes it clear: Sequence A102869, Jonathan Vos Post, Mar 01 2005

#100 ::: Jules ::: (view all by) ::: July 19, 2005, 12:50 PM:

Re: pirates, Alan Hamilton thought the pot would go to the second least senior pirate as the voting carries with half or more, not a majority. With two pirates left, the second to last can propose all the money goes to himself, and the vote will split 50-50.

That's a good start; you've got what happens if you get down to only 2 pirates left right. Now put yourself in the position of the third least senior pirate: how do you avoid the plank, and hopefully net yourself a little cash?

#101 ::: Christopher Davis ::: (view all by) ::: July 19, 2005, 08:35 PM:

Of course, candle's H* puzzle gets thrown off by the closure of Heathrow Terminal 4 to build the extension to Heathrow Terminal 5, and the new routing entailed by that.

#102 ::: Christopher Davis ::: (view all by) ::: July 19, 2005, 10:06 PM:

A sequence puzzle that I don't expect anyone to recognize, though it's findable with a bit of research:

E, SS, TS, CS, NS, F, M, W, B, DS, N, D, E, GS, FH
or
FH, GS, E, D, N, DS, E, S, S, U, NS, CS, TS, SS, E

(Not to be confused with HCKCPW, of course.)

Two questions:

Why is one not the reverse of the other?
and
What connection do these sequences have with MilPhil?

#103 ::: novalis ::: (view all by) ::: July 20, 2005, 04:18 PM:

Melanie, he's playing the odds?

I assume everyone has seen the most excellent two-switch prison problem, but if not, I do recommend it.

#104 ::: Daniel Martin ::: (view all by) ::: July 20, 2005, 10:08 PM:
Which is why I wrote "What comes next?" instead of "what is the next number?" But you're right, and I see your point. Is there another sequence of that form at least as long, which starts with some other number? If not, it's merely a curiosity that I stumbled on by accident, sent to OEIS, and expected to disappear like a feather dropped into the Grand Canyon, with no echo.
I should probably point out that there's a mistake in your sequence, at the number 266124 which should be 1163512 (and this changes subsequent numbers). I discovered this by writing a program to try and find longer all-number sequences.

I will note that the longest all-number sequence only had ten terms, beginning with "11". (Well, aside from the boring sequence beginning with "1")

#105 ::: Melanie S. ::: (view all by) ::: July 21, 2005, 12:27 AM:

Novalis, nope, he's certain (and right, of course).

#106 ::: Jonathan Vos Post ::: (view all by) ::: July 21, 2005, 12:52 AM:

Novalis:

I don't want to spoil Melanie S.'s puzzle, but here's a meta-hint.

Do what humans do best, compared to other creatures. Put yourself in someone else's shoes. Then, from his point of view, imagine him putting himself in someone else's shoes. Consider each alternative. Keep going deeper in this role-playing until you get the answer.

If person A saw situation B, he would do C.
If person A did NOT see situation B, he would do C'.
If person D saw person A do C, he would know E, and so do F.
If person D did NOT see person A do C, he would know E', and so do F'....

Is that clear enough? This also works with Pirates.

#107 ::: Marc G. ::: (view all by) ::: July 21, 2005, 03:19 PM:

Solution to the pirate problem:

The first pirate will take 76 pieces of gold, and pirates 3,5,7,...,49 will take one each.

Explanation: Each pirate needs to offer a better deal to half of the pirates than his successor will offer, but they can offer a worse deal to the other half. So, the 48th pirate can offer 1 piece to the 50th pirate to gain his vote, and therefore the 47th pirate can offer 1 piece to the 49th pirate to gain hers, while refusing to offer the 50th pirate anything. This extrapolates up the line, with each pirate in turn offering opposing halves of the crew 1 piece each.

#108 ::: Jonathan Vos Post ::: (view all by) ::: July 21, 2005, 03:39 PM:

Jules:

Another web site (a Midterm Exam question) axiomatizes the Pirates thus:

A group of perfectly logical pirates finds a treasure chest of 100 coins. The pirates have a captain, a first mate, a second mate, etc. [that is, there is a ranking system that includes everyone.] The Captain proposes a plan for divvying up the gold. The pirates vote on it (ties go to the Captain). If the vote is in the Captains favor, his plan is carried out. If not, the pirates throw the captain overboard and the first mate becomes the Captain, the second mate becomes the first mate, etc., and the new captain proposes a plan for divvying the gold. This continues until one of the plans is agreed upon.

The pirates are perfectly logical and there are three things defining their choice:

i) Self-preservation: If a course of action allows them to live and another will cause them to die, they will always choose the course of action that allows them to live.

j) Greed: Given the choice, they will always choose a course of action that causes them to have the greatest amount of gold.

k) Sanguinity: Given a choice, they will choose a course of action that will cause another pirate to die.

These three things are listed in importance: Self-preservation is always more important than Greed, greed is always more important than sanguinity.

Describe what occurs in each of the following cases:

1) There are 3 pirates

2) There are 30 pirates

3) There are 300 pirates.

[Note, if there are only 2 pirates, the captain could simply make his plan I get all the gold, because there will only be 2 votes, and he will win the tie.]

Here's a blog where people argue over the answer.

I could give other web pages that disagree with each other. I personally consider the definitive analysis of the Pirate's Gold Theorem to be:
Ian Stewart, Scientific American. This is the same wonderful mind who coauthored "Heaven" -- which my son and I considered the novel of the year last year.


#109 ::: Jonathan Vos Post ::: (view all by) ::: July 23, 2005, 02:15 AM:

Time limit, 31.4 seconds. Why is today sometimes called Pi Approximation Day?

#110 ::: Eric Sadoyama ::: (view all by) ::: July 23, 2005, 02:45 AM:

Because it's 22/7?

#111 ::: Jonathan Vos Post ::: (view all by) ::: July 23, 2005, 03:04 AM:

Eric Sadoyama:

Correctomundo! Of course, this was slightly messed up by transforming my posting from Pacific Time to Eastern Standard Tribe.

And, as wikipedia points out:

"The case of Kurt Heegner's work shows that the mathematical establishment is neither infallible, nor unwilling to admit error in assessing 'amateur' work."

By the way, the solution to my puzzle of ::: July 18, 2005, 11:47 PM:

Here's another of my number puzzles which is really about the American English name of the number. What number comes next?

2, 0, 11, 14, 102, 101, 111, 114, 1014, 1102, 1101, 1111, 1014, 11114, 14114, 101111, 101114, 111114 is:

Smallest nonnegative integer whose standard American English name has n vowels.

a(1) = 2 [twO], where "two" is the smallest integer with 1 vowel.
a(2) = 0 [zErO], where "zero" is the smallest integer with 2 vowels.
a(3) = 11 [ElEvEn]
a(4) = 14 [fOUrtEEn]
a(5) = 102 [OnE hUndrEd twO]
a(6) = 101 [OnE hUndrEd OnE]
a(7) = 111 [OnE hUndrEd ElEvEn]
a(8) = 114 [OnE hUndrEd fOUrtEEn]
a(9) = 1014 [OnE thOUsAnd fOUrtEEn]
a(10) = 1102 [OnE thOUsAnd OnE hUndrEd twO]
a(11) = 1101 [OnE thOUsAnd OnE hUndrEd OnE]
a(12) = 1111 [OnE thOUsAnd OnE hUndrEd ElEvEn]
a(13) = 1014 [OnE thOUsAnd fOUrtEEn]
a(14) = 11114 [ElEvEn thOUsAnd OnE hUndrEd fOUrtEEn]
a(15) = 14114 [fOUrtEEn thOUsAnd OnE hUndrEd fOUrtEEn]
a(16) = 101111 [OnE hUndrEd OnE thOUsAnd OnE hUndrEd ElEvEn]
a(17) = 101114 [OnE hUndrEd OnE thOUsAnd OnE hUndrEd fOUrtEEn]
a(18) = 111114 [OnE hUndrEd ElEvEn thOUsAnd OnE hUndrEd fOUrtEEn]
a(19) = 114114 [OnE hUndrEd fOUrtEEn thOUsAnd OnE hUndrEd fOUrtEEn]


#112 ::: Jules ::: (view all by) ::: July 23, 2005, 10:56 AM:

I personally consider the definitive analysis of the Pirate's Gold Theorem to be:
Ian Stewart, Scientific American.

Thank you. That was where I originally found the puzzle and its analysis, although I had forgotten where it was.

This is the same wonderful mind who coauthored "Heaven" -- which my son and I considered the novel of the year last year.

I have yet to read it, although I did read Wheelers, also by Profs Stewart & Cohen, and found it to be a very intriguing book. We don't seem to get many good new first contact stories these days, but that was certainly enjoyable. The aliens were unforgettable.

#113 ::: Jonathan Vos Post ::: (view all by) ::: July 23, 2005, 11:14 AM:

You're welcome, Jules.

Sometimes there's a nice sequence of integers that just peters out. For example, today this appeared (I submitted it 20 July 2005) on the Online Encyclopedia of Integer Sequences: A108942:

0, 1, 2, 3, 4, 5, 6, 7, 9, 18, 33

Numbers n such that 10^n is the product of two integers without any zero digits.

The first number must be some power of 2, the second the same power of 5. Such a pair is the only positive solution, because neither factor could have both a 2 and a 5 in its prime factorization, or else it would be a multiple of 10 and would thus have a 0 as its last digit, which is ruled out. "We may wonder what powers of 10 are products of two integers without any zero digits. For large numbers, this is very unlikely because there will normally be 10% of zeroes among many random digits... In fact, there seems to be only 11 possibilities... The probability is roughly (0.9)^n that the nth power of 10 would yield a solution. So, the expected number of solutions above the nth power of 10 is someting like 10*(0.9)^n. Since we've actually checked that there's no other solution below n = 1500, we can be very confident that we've not missed anything..."
G. P. Michon, What two integers without zero digits have a product of 1000000000?

10^0 = 1 * 1
10^1 = 2 * 5
10^2 = 4 * 25
10^3 = 8 * 125
10^4 = 16 * 625
10^5 = 32 * 3125
10^6 = 64 * 15625
10^7 = 128 * 78125
10^9 = 512 * 1953125
10^18 = 262144 * 3814697265625
10^33 = 8589934592 * 116415321826934814453125

And that's all she wrote...

But note that it starts out 0, 1, 2, 3, 4, 5, 6, 7, and then changes. This is a warning against assuming without proof that a pattern will continue indefinitely. This is a mistake associated with "The Law of Small Numbers".

I like this summary: "There aren't enough small numbers to meet the many demands made of them."

#114 ::: Jonathan Vos Post ::: (view all by) ::: July 23, 2005, 06:16 PM:

I'm presuming that Teresa started this A Small Puzzle" thread because such puzzles are of interest, not out pure geekiness, not out of overweening faith in Logic, but because Teresa has a feeling for the aesthetics of Mathematics, consistent with her belief in the beauty of Fibonacci sequences in the arrangement of flower petals. With that in mind, here's a nice quotation from Wikipedia on mathematics: "Inspiration, aesthetics and pure and applied mathematics."

"... With the increase in our mathematical knowledge, mathematics itself has become a source of inspiration. Mathematics is inspiring to mathematicians because it has some intrinsic aesthetics or inner beauty, which is hard to explain. Mathematicians value especially simplicity and generality and when these seemingly incompatible properties combine in a piece of mathematics, as in a unifying generalization for several subfields, or in a helpful tool for common calculations, often that piece of mathematics is called beautiful. Since the result of mathematics inspired by mathematics is often pure mathematics and thus has no applications outside of mathematics yet, the only value it has is in its aesthetics. Surprisingly often, it has happened that pure mathematics, which was considered only of interest to mathematicians, has become applied mathematics because of some new insight, as if it anticipated later needs...."

By the way, Today's featured article on Wikipedia is:

"The Monty Hall problem is a puzzle in probability that is loosely based on the American game show Let's Make a Deal. The name comes from the show's host Monty Hall. In this puzzle a player is shown three closed doors; behind one is a car, and behind each of the other two is a goat. The player is allowed to open one door, and will win whatever is behind the door. However, after the player selects a door but before opening it, the game host opens another door revealing a goat. The host then offers the player an option to switch to the other closed door. Does switching improve the player's chance of winning the car? The answer is yes switching results in a 2/3 chance of winning the car. The problem is also called the Monty Hall paradox, in the sense that the solution is counterintuitive, although the problem is not a logical self-contradiction."

#115 ::: Byna ::: (view all by) ::: July 28, 2005, 12:30 PM:

Is there any generic formula or theory to predict "what comes next"?

#116 ::: Jonathan Vos Post ::: (view all by) ::: July 28, 2005, 02:37 PM:

Byna:

In general, no. Sometimes, nobody CAN know, even in theory. Nobody meaning any kind of finite mental being that we can imagine. Paul Erdos, the most-published Mathematician of our age, assumed that God (whom he considered the ultimate dictator) knew all solutions to all equations. Erdos referred to particularly elegant proofs as "from the Book" -- meaning God's book. Today, of course, one has to imagine God's Blog.

#117 ::: Deepexplorer ::: (view all by) ::: July 30, 2005, 05:18 PM:

If the question is 'what integer comes next' in an integer sequence, Markov chains can be used to find the recurrence of the distance between successive integers of that sequence. I have seen it in the context of data prefetching in computer architecture. But I am not fully aware of the Markov chains algorithm.

#118 ::: Gene Klimov ::: (view all by) ::: June 05, 2008, 11:30 AM:

When I searched my publications ( using ‘google’ with G. P. Klimov) in your site I saw a remark #11 about ‘annoying problem, which is attributed to G. P. Klimov in my math reasoning book’. In all solutions of this problem the “..mail carriers..” was ignored. It means they are not retired and the age

#119 ::: Xopher ::: (view all by) ::: June 05, 2008, 12:57 PM:

Gene, please go on. It looks like your comment was truncated. Are you saying the age of the other mail carrier must be less than 65? I can't think that the red hair means anything, except to indicate a "middle son" exists, which means no two of the sons are the same age.

I don't see how the windows thing helps at all, since a building could have any arbitrary number of windows from zero (e.g. a telephone exchange) to thousands (e.g. One World Trade Center of late lamented memory); the range is confined more than that by the fact that the mail carrier must be at least 12 to 13 years older than his oldest son, and must be under 65 (if that's what you were going to say).

So we have:

s1 × s2 × s3 < 65;
s1 < 52;
s1 > s2 > s3
I still can't solve this without just trying numbers.

#120 ::: Gene Klimov ::: (view all by) ::: June 05, 2008, 03:36 PM:

When I searched my publications ( using ‘google’ with G. P. Klimov)
in this site I saw your remark #11 about ‘annoying problem, which is attributed to G. P. Klimov in my math reasoning book’.

[1] I did not published this my problem, I have shown it to some my friends only.
[2] In all solutions of this problem (in the site) the “..mail carriers..” was ignored. It means they are not retired and the “age” less 72. (there is another solution (3,3,8) for “age”=72).
Xopher, thank you very much.

#121 ::: Xopher ::: (view all by) ::: June 05, 2008, 04:03 PM:

You're welcome. I didn't know what age allows mail carriers to retire.

But surely (3,3,8) doesn't work? Which of the twin three-year-olds is the "middle son"?

#122 ::: Gene Klimov ::: (view all by) ::: June 05, 2008, 04:26 PM:

2*6*6=72, 2+6+6=14,
3*3*8=72, 3+3+8=14,
The oldest one has red hair,
So (3,3,8) is solution.

#123 ::: Gene Klimov ::: (view all by) ::: June 05, 2008, 06:20 PM:

In #11
“My middle son is red-haired.”
should be changed by
“My oldest son is red-haired.”

#124 ::: Debbie sees spam ::: (view all by) ::: January 09, 2012, 03:17 AM:

Link spam

#125 ::: Stefan Jones suspects spam ::: (view all by) ::: March 09, 2012, 07:50 PM:

IT service spice meat product.

#127 ::: Akira ::: (view all by) ::: September 13, 2012, 02:09 AM:

Can someone help me solve this one asap

4, 13, 22, 34, 49 ... next number

If you know just post I need it for a test

#128 ::: Akira ::: (view all by) ::: September 13, 2012, 02:09 AM:

Can someone help me solve this one asap

4, 13, 22, 34, 49 ... next number

If you know just post I need it for a test

#129 ::: Jim Macdonald ::: (view all by) ::: September 13, 2012, 02:24 AM:

Welcome to Making Light, Akira.

We don't usually offer Homework Help here. Are you constructing the test or taking it?

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