Help me solve this Douglas Coupland puzzle.

May 27, 2006 6:54 PM

I am reading Douglas Coupland's new book (JPod) and I'm always a little frustrated by him. He seems to get the general vibe of nerds but is horribly off on details sometimes (in the first 50 pages, he makes reference to a "56k floppy disk"), which is agonizing. Anyway, he posits a math problem (I don't think this counts as a spoiler, I'll even omit the context and just pose the problem with page number, but if you really don't want to see anything about the book, don't read on, I guess.)

"Anyway, send me an email or even phone me. It's area code 604, and the number itself is a seven-digit prime which, when squared, is two digits short of being a factorial." p.50, hardcover edition.

Not having had much call to deal with primes in the last five years or so, I went to mathematica. Apparently the 78,499th through 664,579th primes are seven-digit, so no love there. So I went to a factorial table. Factorials of course, quickly grow many zeroes in their tails. This is a problem since the square of any number with trailing zeroes is nonintegral if it has an odd number of trailing zeroes and is a power of ten, if it has an even nonzero number of trailing zeroes.

Since any seven-digit number squared is 13-14 digits, I figured I'd look at ones with 13-15 digits in their non-trailing zeroes. That way I could drop all the zeroes as a redundant digit, plus one. I included 21! because its least significant non-trailing zero digits are redundant.

18! = 6402373705728000
19! = 121645100408832000
20! = 2432902008176640000
21! = 51090942171709440000

None of these is the square of a prime.

Ideas?

posted by oxonium to Science & Nature (42 answers total)

Response by poster: By the way, 604 is a BC, Canada area code, so be mindful of that if you decide to call it. Here's hoping Coupland didn't just make this up because it "sounded Googley."
posted by oxonium at 6:56 PM on May 27, 2006

Squaring isn't taking the square root, it's multiplying by itself.


I never got past the whole preachiness in Girlfriend in a Coma myself.
posted by fvw at 7:18 PM on May 27, 2006

Response by poster: Right, I'm talking about taking the square root of the factorial (i.e., the factorial contains the prime squared somewhere).

I could never get through GenX, life without god, etc., but I love microserfs and still read it from time to time. This looks promising but frustrating as usual.
posted by oxonium at 7:21 PM on May 27, 2006

Well, two digits short of a prime. So maybe that means any two digits could be changed to turn it into the square of a prime. At 14 digits, that yeild 14*13 = 182 possible numbers. Or 364 if you going to try it with 19! and 20!

Also, obviously, if a number is a factorial, it can't be prime After all the fractorial of X is composed of all the factors < x. br>
I doubt, but can't prove at the moment that a fractorial can be the square of a prime either.

So it's probably just a joke or some mathimatical gobltygook.
posted by delmoi at 7:26 PM on May 27, 2006

N! can't possibly be the square of prime number P. P^2 can be factored into P*P; N! has factors N, N-1, N-2, etc., none of which are repeated.

What do you mean by "nonintegral"?
posted by Wet Spot at 7:26 PM on May 27, 2006

Oh, right, I was slightly thrown by a missing "root" before "of any number with trailing zeroes", I get what you're doing now.
posted by fvw at 7:27 PM on May 27, 2006

Response by poster: By nonintegral I just mean "not an integer," sorry, I'm working from the phys. chem. literature when I think about math, so the words are probably different. As for "factorial can't be prime," look again at what I'm doing. He's talking about "two digits short," so presumably something is lopped off (end digits, maybe more than one zero). I don't think these are precluded from being prime, but I am not so hot with this stuff anymore.
posted by oxonium at 7:35 PM on May 27, 2006

I was thinking two digits away from a factorial meant that sqrt(n!/100)=y, where y is the 7-digit phone number.
posted by evariste at 7:38 PM on May 27, 2006

Response by poster: Evariste: yeah, that was my first thought, but there are no 15-16-digit factorials with two and only two trailing zeroes.
posted by oxonium at 7:39 PM on May 27, 2006

::after too long playing with the calculator::

I think I would just email the guy.
posted by evariste at 7:53 PM on May 27, 2006

Response by poster: Thanks for giving it a go.
posted by oxonium at 7:56 PM on May 27, 2006

Let's start with "...two digits short of being a factorial". The "two digits" is kind of awkward phrasing. Let's assume it means that n^2 + 2 = m!. Any factorial larger than 4! will end in a zero. Any number ending in zero minus 2 will result in a number ending in 8. There is no number multiplied by itself that gives a number ending in 8. Therefore, there is no solution for n^2 + 2 = m!.

Now let's assume it means there are missing digits on the square of the prime. In a factorial, for every factor that is a multiple of five, the result will have another zero on the end. Therefore, any factorial greater than 14! will have at least 3 zeroes. Dropping two of those leaves another zero and no prime squared will result in a number ending in zero.

14! with the last two digits dropped is only a 7-digit number, so it can't be the solution. Either I'm not understanding the problem, or Coupland is still missing the true Essence d'Nerd.
posted by forrest at 8:00 PM on May 27, 2006

I hope someone solves it. I would be infuriated if it was just a plot device with no solution.
posted by evariste at 8:00 PM on May 27, 2006

I didn't find the answer (I think it's either no solution or there's some meaning to being two digits short that we're not getting), but I did find tons of stupid information about seven digit primes. (Link is slow to load.)
posted by Xalf at 8:06 PM on May 27, 2006

Response by poster: Forrest: Yeah, I dealt with both of those ideas too. The only thing I can think of is treating redundant least-significant digits as one "digit." It is terribly sloppy (since 0000 is four digits, of course), but the only hope it has of being solved. To wit:

Dropping the two least-significant "digits" by my sloppy definition of, say:

20! = 2432902008176640000

Gives 24329020081766 (4 redundant zeroes and 1 nonredundant four)

21! = 51090942171709440000

Gives 51090942171709 (4 redundant zeroes and 2 redundant fours)

Lopping off digits from the beginning is simply not an option; you end up with a power of ten in the factor and/or sqrt(10) which will give a solution which is not an integer.

An idea I haven't yet tried is removing ALL the zeroes from the number (i.e., including interstitial zeroes) plus a digit. Again, any solution that relies on removing multiple digits and calling them one digit is technically incorrect, but this seems to be the only possibility.
posted by oxonium at 8:07 PM on May 27, 2006

Response by poster: Thanks for helping xalf, anyone who is still working, note that his link is not an all-inclusive link of seven-digit primes. Mathematica gives:

In[16]:=
Prime[78499]

Out[16]=
1000003

In[40]:=
Prime[664579]

Out[40]=
9999991

Meaning the 78,499th and 664,579th primes, these bracket all seven-digit primes.
posted by oxonium at 8:11 PM on May 27, 2006

A lot of the items on that list aren't primes at all, but links to certain properties of primes. For instance, 1000000 is obviously not prime.
posted by evariste at 8:14 PM on May 27, 2006

Assuming this is a real puzzle...

Working at the problem from the other end, I looked for D Coupland in the 604 area code. There's one, but the number isn't prime, so that's not it.

Unless there's a trick in the solution, I think forrest is right. Any factorial in the right range is going to have more than two trailing zeros, and primes don't have any.
posted by justkevin at 8:33 PM on May 27, 2006

List of all valid telephone prefixes for area code 604 (Push the List All button).

Anyone want to see if that makes the list of 7 digit prime numbers more manageable?
posted by ceribus peribus at 8:33 PM on May 27, 2006

Response by poster: I am off for the night. It embiggens my spirit to see people giving it a go, will check back tomorrow, thanks!
posted by oxonium at 8:39 PM on May 27, 2006

On second look, that's 540 exchanges. Which is better than 999, but not by much. I blame Vancouver.
posted by ceribus peribus at 8:52 PM on May 27, 2006

Another possibility for the two digits: what if it means subtract two single digits?
posted by smackfu at 8:55 PM on May 27, 2006

Best answer: I'm starting to think we're trying to get Voynich's phone number, but I considered if by "two digits short" he meant it in the Wheel of Fortune sense-- that you could remove all instances of any two digits. One would obviously have to be zero, but the other could be anything.

Short answer: no, there are no two digits where removing every instance from a factorial will give a solution. (only first couple will give a square at all) Checked up to 100!
posted by justkevin at 9:14 PM on May 27, 2006

Or, perhaps it's that 2 digits are different from the sqrt(m!)? In which case, you know the last digit must be one of them, and there's only a limited number of choices for that digit. Then, it's an ugly problem of comparing the factorials digitwise to primes.

Sounds like fun for someone else.
posted by JMOZ at 9:18 PM on May 27, 2006

Best answer: OK, make my last comment that n^2 differs by 2 digits from m!. In that case, you're narrowed down to m= 15 or 16. Of course, that leaves you with 13-14 digits that can be changed to get a perfect square, which is quite a list of numbers to try, but might not be too hard to program in Mathematica. (I don't have mathematica at home, or I'd consider trying it.)

There's only a few digits a perfect square can end with (and if it's the square of a prime, 0 isn't one of them), so that makes it a lot easier; you know the last digit has to be one of the ones which changes.
posted by JMOZ at 9:37 PM on May 27, 2006

Best answer: There's only a few digits a perfect square can end with (and if it's the square of a prime, 0 isn't one of them), so that makes it a lot easier; you know the last digit has to be one of the ones which changes.

Squares of primes greater than 5 end in 1 or 9.
posted by justkevin at 9:48 PM on May 27, 2006

FYI: Doug is likely reading this or will at some point in the near future. /derail
posted by shoepal at 11:07 PM on May 27, 2006

Best answer: JMOZ, it is possible to generalize your idea to use Levenshtein distance (i.e. edit distance or the number of deletions, insertions or substitutions). Unfortunately, the closest square of a seven digit prime to a factorial has a Levenshtein distance of four, not two:

prime: 2965601
square: 8794789291201
factorial: 14! or 87178291200
distance: 4
posted by RichardP at 2:14 AM on May 28, 2006

piggy-backing on ceribus peribus's post: I would start with the North Vancouver exchanges, and then broaden out. (I have no idea if Coupland currently lives in North Vancouver, but he is from the North Shore). From the previous link, the North Vancouver exchanges are 903, 904, 924, 929, 980, 981, 982, 983, 984, 985, 986, 987, 988, 989, 990, 998.
posted by crazycanuck at 4:39 AM on May 28, 2006

Douglas Coupland isn't worth this much trouble.
posted by cropshy at 5:34 AM on May 28, 2006

crazycanuck: Coupland lives in West Vancouver - British Properties, I think. If that helps.
posted by meringue at 8:22 AM on May 28, 2006

From that list, narrow it down to these numbers:

921, 922, 923, 925, 926, 963

On the other hand, if he's given his cell number, the area code is probably different.
posted by meringue at 8:26 AM on May 28, 2006

For a cell number, the area code would be 778.
posted by meringue at 8:29 AM on May 28, 2006

Even though Coupland doesn't live in North Vancouver anymore, I'd say his fascination with the area would be enough to make it a North Van number.

However, I think you should consider other possibilities.
604 254-7191 < --- his local press contact, for examplebr> I'm not a mathematician, so I can't tell you if that number works out.

Meringue: not all cell phones are 778. I got a new cell phone and number last week and I'm still 604.
posted by acoutu at 9:38 AM on May 28, 2006

Look at the factorials. Any that would be relevant are even or start with even numbers when you remove the end or beginning digits. All those zeroes, you know?

So he's got to mean add 2. As in +2. At least, I think so.
posted by acoutu at 9:52 AM on May 28, 2006

I gave up bothering with Coupland maths questions after the one in Microserfs. I don't have a copy here, but it's apparently one that kept his programmer characters off their work all day. I mentioned it to a programmer friend of mine and he had the answer in about 5 seconds.
posted by bonaldi at 10:13 AM on May 28, 2006

Whoops - I should have gone back and re-read the question. He says it's 604 in the book, apparently. My bad. As for the math part, I'm completely hopeless there. I just like Coupland & his approach to being an artist, so the idea of finding out his phone number and being able to drunk-dial him at 3 am is appealing in some weird way. NOT that I would ever do that! (Though I'm sure that that's the reason he got someone to think up a tricky math problem). So Doug, if you're reading this, I promise never to call you, 3 am or no, even if I ever find out your phone number. ;)
posted by meringue at 11:22 AM on May 28, 2006

Response by poster: So, it looks like we've a) tried removing just two trailing digits (single instances), which can't work for factorials of the appropriate size because they all have 10 as a factor, which precludes prime integer square root, b) tried removing two trailing digits (multiple instances), which didn't work, and c) tried removing all instances of two interstitial and trailing digits (multiple instances, single instances would have the trailing zero problem). None of these work. I can't think of another way to approach it which is even close to technically correct.

Anyone?
posted by oxonium at 11:53 AM on May 28, 2006

Best answer: oxonium, as I mentioned above, I also tried all possible combinations of deletions, insertions or substitutions. The minimum number of such edits required to transform a seven digit square into a factorial is four (of which there is only one, which I provided earlier).
posted by RichardP at 12:17 PM on May 28, 2006

um, for "a seven digit square" read "square of a seven digit prime."
posted by RichardP at 12:20 PM on May 28, 2006

Response by poster: Oop, missed yours, thanks. I hadn't thought of it as edit distance, I was thinking only of addition/deletion.
posted by oxonium at 12:36 PM on May 28, 2006

Wait - there's a math riddle in Microserfs? I'm a math-challenged geek, and if there's such a puzzle, it flew over my head.

Where is it?
posted by awenner at 4:23 AM on June 7, 2006