# A math puzzle to try your hand at: **URL:** \ **Category:** Miscellaneous and Personal Stuff I Must Share\ **Created:** [June 22, 2005, 12:41pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346 "2005-06-22T12:41:31Z")\ **Posts on this page:** 18\ **Page:** 1
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [June 22, 2005, 12:41pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/1 "2005-06-22T12:41:31Z")
I thought of this and thought I’d see how quickly some of the bright dopers we have here can solve it. It’s a new way of writing numbers, a little like binary except more confusing and impractical. 😉 I’ll list out the first twenty-some numbers in increasing order - if you need more information, just ask questions and I’ll do my best to answer without giving the secret away. If you think you’ve figured it out, try posting the number seventy-five in the same format as I’m using. Here’s the list to get you started. If I made any mistakes here, I’m sorry, I tried my best to work them all out right. 0 1 10 2 100 11 1000 3 20 101 1000 12 10000 1001 110 4 100000 21 1000000 102 1010 1001 10000000 13 200 10001 30 1002 ---
**Author:** ![wolfstu](https://avatars.discourse-cdn.com/v4/letter/w/d6d6ee/32.png) [@wolfstu](https://boards.straightdope.com/u/wolfstu)\ **Post date:** [June 22, 2005, 12:49pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/2 "2005-06-22T12:49:35Z")
> [@chrisk](#): > > I’ll list out the first twenty-some numbers in increasing order … > > 0 > … > 1000 > … > 1000 > … I thought there was a pattern, at first, but then two different numbers were represented by the same symbol (“1000”), which, to my mind, is either a typo or a pretty serious flaw. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [June 22, 2005, 12:54pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/3 "2005-06-22T12:54:26Z")
> [@wolfstu](#): > > I thought there was a pattern, at first, but then two different numbers were represented by the same symbol (“1000”), which, to my mind, is either a typo or a pretty serious flaw. Okay, yes, mistake, I skipped a ‘place’ of positional notation for every number that should have been at least five digits long. Try that again: 0 1 10 2 100 11 1000 3 20 101 10000 12 100000 1001 110 4 1000000 21 10000000 102 1010 10001 100000000 13 200 100001 30 1002 ---
**Author:** ![lno](https://avatars.discourse-cdn.com/v4/letter/l/a8b319/32.png) [@lno](https://boards.straightdope.com/u/lno)\ **Post date:** [June 22, 2005, 2:46pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/4 "2005-06-22T14:46:19Z")
I’m only halfway there but I’ve got to get back to work, so here’s some information for others to use. The first thing I noticed is that the powers of two are related. That is: 2^1 = 1 2^2 = 4 2^3 = 8 Etc, so your value of “3” is our base-10 8. I would extrapolate your value of “65536” to be our 16. The next is that the powers of three are also related. 3^1 = 1 (but 10 for you) 3^2 = 9 ( but 20 for you) 3^3 = 27 (but 30 for you) I would predict that 3^4, 81, would be written as “40” in your system. Following that, your primes are in a predictable pattern: 3 = 10 5 = 100 7 = 1000 11 = 10000 13 = 100000 17 = 1000000 19 = 10000000 23 = 100000000 I surmise that 29 would be written as “1000000000”, with each subsequent prime getting an additional zero. Then there’s some duplication activity going on: 7 = 1000 14 = 1001 28 = 1002 Following that pattern, 56 would be “1003”. Each doubling seems to increment the final digit in the number. Given that 75 is not a power of two or three, and is not prime, this may be a more productive line of thought. Looking at your multiples of five, I see: 5 = 100 10 = 101 20 = 102 So with the doubling seen above, 40 would be “103”. However, we still can’t get to 75 directly from there. Maybe look at powers again…? 5^1 = 100 5^2, or 25 = 200 I’d predict 5^3, 125, to be “300”. Still not directly useful. Solving for 75 is left as an exercise for the reader. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [June 22, 2005, 3:08pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/5 "2005-06-22T15:08:19Z")
> [@lno](#): > > 2^3 = 8 > Etc, so your value of “3” is our base-10 8. I would extrapolate your value of “65536” to be our 16. > > 3^3 = 27 (but 30 for you) > I would predict that 3^4, 81, would be written as “40” in your system. > > 23 = 100000000 > I surmise that 29 would be written as “1000000000”, with each subsequent prime getting an additional zero. > > 28 = 1002 > Following that pattern, 56 would be “1003”. Each doubling seems to increment the final digit in the number. Given that 75 is not a power of two or three, and is not prime, this may be a more productive line of thought. > > 20 = 102 > So with the doubling seen above, 40 would be “103”. However, we still can’t get to 75 directly from there. Maybe look at powers again…? > > 5^2, or 25 = 200 > I’d predict 5^3, 125, to be “300”. Still not directly useful. > > Solving for 75 is left as an exercise for the reader. Very well done lno! I’d say you’re more than halfway there - there’s probably just one more step that would break the entire puzzle wide open. All of your extrapolations, 65536, 81, 29, 56, 40, and 125 are correct. Except that 65536, to follow the pattern would have to be represented with a single digit that represents our number 16, so that it can fit into the first notational position. (sixteen written out one-six would actually be the representation for 196 – there’s another clue, maybe.) Among the several limitations of this scheme as a real numbering system are the need for arbitrarily many distinct digits to represent sufficiently high powers of 2 (and 3, and so on,) and the fact that relatively low primes need so many digits to convey them. But it’s a fun game anyway, I think. 🙂 ---
**Author:** ![amarinth](https://avatars.discourse-cdn.com/v4/letter/a/3be4f8/32.png) [@amarinth](https://boards.straightdope.com/u/amarinth)\ **Post date:** [June 22, 2005, 4:33pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/6 "2005-06-22T16:33:44Z")
> [@chrisk](#): > > Among the several limitations of this scheme as a real numbering system are the need for arbitrarily many distinct digits to represent sufficiently high powers of 2 (and 3, and so on,) and the fact that relatively low primes need so many digits to convey them. But it’s a fun game anyway, I think. 🙂 Well, thanks to **lno** , I realized you’d started with 1, and not with 0. (which completely threw me off. cute. of course, not being able to distinguish between 30, 10240, 354294, and 2 and a half decillion might be considered a problem for some people. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [June 22, 2005, 5:32pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/7 "2005-06-22T17:32:26Z")
> [@amarinth](#): > > Well, thanks to **lno** , I realized you’d started with 1, and not with 0. (which completely threw me off. > > cute. of course, not being able to distinguish between 30, 10240, 354294, and 2 and a half decillion might be considered a problem for some people. Just for the record… did you figure out anything besides that I started with 1? If so, care to try solving the 75 problem?? As far as distiguishing between numbers like that, I imagine it’d be a question of training yourself to ‘read’ numbers in this wonky alphabet. Wouldn’t be hard to get into the right ballpark, though transcribing quickly might be a lot to ask for. ---
**Author:** ![amarinth](https://avatars.discourse-cdn.com/v4/letter/a/3be4f8/32.png) [@amarinth](https://boards.straightdope.com/u/amarinth)\ **Post date:** [June 22, 2005, 7:42pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/8 "2005-06-22T19:42:08Z")
Of course, I can solve for 75. (I solved for 30, 10240, 354294, and 2.5 decillion. Did you?) 75 would be 210. Those other four numbers could all arguably be written as 111. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [June 22, 2005, 9:09pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/9 "2005-06-22T21:09:47Z")
[spoiler]Whoops, I didn’t get that, sorry. I was at work and didn’t stop to try solving all of those. I don’t agree that they could all be represented as 111, but only because I didn’t see the usual rules of base-10 positional notation applying to this language. 30 would be 111. If we adopt A as an eleventh digit, after the style of hexadecimal, then 10240 would be 10A , and 354294 would be A1 Your particular number in the vicinity of 2.5 decillion would require a digit number 111, which seems awfully unwieldy… but then, just when do you need to refer to numbers in the vicinity of 2.5 decillion anyway?? Strictly speaking, 2.5 decillion, exactly 2.5 \* 10^33 , would be represented as X0V if we continue using letters in place of digits up all the way to Z as 36. X would be the 34th digit, V would be the 32nd.[/spoiler] Why yes, I’m a math geek, how could you tell? 😃 ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [June 22, 2005, 9:37pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/10 "2005-06-22T21:37:07Z")
correction: [spoiler]If we adopt B as an eleventh digit, after the style of hexadecimal, then 10240 would be 10B, and 354294 would be B1 (A is the tenth digit, after 9) 2.5 \* 10^33 , would be represented as Y0W if we continue using letters in place of digits up all the way to Z as 35. Y would be the 34th digit, W would be the 32nd. [/spoiler] I think… 🙂 ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [June 24, 2005, 12:44am UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/11 "2005-06-24T00:44:42Z")
Okay, I’m using my one bump. Any other smart math people want to try this?? ---
**Author:** ![TJdude825](https://avatars.discourse-cdn.com/v4/letter/t/dec6dc/32.png) [@TJdude825](https://boards.straightdope.com/u/TJdude825)\ **Post date:** [June 24, 2005, 3:50am UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/12 "2005-06-24T03:50:29Z")
First of all, I believe it makes it much easier to make this table before starting: 1 - 0 2 - 1 3 - 10 4 - 2 5 - 100 6 - 11 7 - 1000 8 - 3 9 - 20 10 - 101 11 - 10000 12 - 12 13 - 100000 14 - 1001 15 - 110 16 - 4 17 - 1000000 18 - 21 19 - 10000000 20 - 102 21 - 1010 22 - 10001 23 - 100000000 24 - 13 25 - 200 26 - 100001 27 - 30 28 - 1002 Now. **Ino** ’s post gives us all primes (1, 10, 100, 1000, 10000, etc.) all powers of primes (1000, 2000, 3000, etc.) and all numbers expressible as (2^n)\*p where n is an integer and p is a prime (10000, 10001, 10002, 10003, etc.) I think the next step is to figure out the pattern for all multiples of a given prime. The known multiples of primes are: Twos: 2 - 1 4 - 2 6 - 11 8 - 3 10 - 101 12 - 12 14 - 1001 16 - 4 18 - 21 20 - 102 22 - 10001 24 - 13 26 - 100001 28 - 1002 Threes: 3 - 10 6 - 11 9 - 20 12 - 12 15 - 110 18 - 21 21 - 1010 24 - 13 27 - 30 Fives: 5 - 100 10 - 101 15 - 110 20 - 102 25 - 200 Sevens: 7 - 1000 14 - 1001 21 - 1010 28 - 1002 Elevens: 11 - 10000 22 - 10001 Thirteens: 13 - 100000 26 - 100001 Note that a number composed of two primes must fit two patterns (i.e. 15 fits the 3-pattern and the 5-pattern). After the prime itself, replace the final zero with a 1 to get the next number (2p), with the exception of 4=2\*2, expressed as “2” because it’s a power. Beyond that, I feel like I can kind of “see” the pattern, but I don’t know exactly what it is. Hopefully all my tables will help others. Meanwhile, I’m still working on this. ---
**Author:** ![TJdude825](https://avatars.discourse-cdn.com/v4/letter/t/dec6dc/32.png) [@TJdude825](https://boards.straightdope.com/u/TJdude825)\ **Post date:** [June 24, 2005, 4:09am UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/13 "2005-06-24T04:09:41Z")
Okay, wait! I think I got it! [spoiler]Write the number as a “prime factorization” meaning divide it all the way out into its prime factors. (e.g. 75 = 3_5_5 or 3\*5[sup]2[/sup]. The furthest place to the right is the number of twos in the factorization (or the exponent next to the two… in this case, zero). The 2nd place from the right (what looks like the 10’s place) is the number of threes (in this case, 1). The next place is the number of 5’s (in this case, 2) and so on. Any number that’s a multiple of 2^10, 3^10, 5^10, etc. requires new digits beyond 9. I believe all of **Ino** ’s observations come about naturally as a result of the system. 75 is “210” There are definite advantages to this system, for example, reducing fractions. Actually, that might be the ONLY advantage. That was fun. [/spoiler] Right? ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [June 24, 2005, 10:22am UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/14 "2005-06-24T10:22:16Z")
There’s a serious disadvantage to the scheme, which is that it needs an infinite amount of notation. Our usual notation only requires 10 different digits. If 2^n is represented by the nth character in the notation, then you need a new character for every power of 2. You might overcome hat by some means like using ordinary decimal notration within parentheses, e.g., represent 1024 as (10), 2048 as (11), 3072 as 1(10), etc., but that would spoil the purity of the system. But, apart from that flaw, the system is pretty neat! ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [June 24, 2005, 4:52pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/15 "2005-06-24T16:52:50Z")
> [@Giles](#): > > There’s a serious disadvantage to the scheme, which is that it needs an infinite amount of notation. Our usual notation only requires 10 different digits. If 2^n is represented by the nth character in the notation, then you need a new character for every power of 2. You might overcome hat by some means like using ordinary decimal notration within parentheses, e.g., represent 1024 as (10), 2048 as (11), 3072 as 1(10), etc., but that would spoil the purity of the system. > > But, apart from that flaw, the system is pretty neat! Yeah, that’s pretty much my thoughts about it too - nice idea with the parenthetical notation. 🙂 and tjdude is right too of course. 🙂 ---
**Author:** ![TJdude825](https://avatars.discourse-cdn.com/v4/letter/t/dec6dc/32.png) [@TJdude825](https://boards.straightdope.com/u/TJdude825)\ **Post date:** [June 24, 2005, 4:54pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/16 "2005-06-24T16:54:11Z")
I agree that it’s a flaw, but if you use A for 10 and B for 11 and so on, then that gets you all the way to 35. Then you’d probably start using lowercase letters or Greek letters and it would get to the point where there were unwritable numbers, but they’d rarely come up. Also, I’m not sure how you’d represent non-integers. One half could be 0/1, nine tenths could be 20/101, but I don’t know how you’d do the equivalent of “decimals” (like 0.5 or 0.9) For that matter, I don’t see any way of writing the number zero. Maybe just “00” by definition? ---
**Author:** ![TJdude825](https://avatars.discourse-cdn.com/v4/letter/t/dec6dc/32.png) [@TJdude825](https://boards.straightdope.com/u/TJdude825)\ **Post date:** [June 24, 2005, 10:33pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/17 "2005-06-24T22:33:51Z")
A couple more things. If we used this, discovering new primes would be a no-brainer. You’d see news articles like “After last week’s discovery of the 210th prime, 1000000000000000000000000000000000000000000000000000000000000000000000000000, mathematicians have just used an exhaustive computer search to discover the 10000010th prime, 1000000000000000000000000000000000000000000000000000000000000000000000000000. However, it is not known just how big this number is.” I thought this was interesting: 12 - 12 Any other numbers represented the same in ordinary decimal numerals and in **chrisk** erals? Can it be proven that there are none, or that there an infinite number of them? ---
**Author:** ![Hal\_Briston](https://avatars.discourse-cdn.com/v4/letter/h/a9a28c/32.png) [@Hal\_Briston](https://boards.straightdope.com/u/Hal_Briston)\ **Post date:** [June 24, 2005, 10:36pm UTC](https://boards.straightdope.com/t/a-math-puzzle-to-try-your-hand-at/309346/18 "2005-06-24T22:36:29Z")
> [@chrisk](#): > > Very well done lno! If only he were that good at making football picks. \ 😃