A math puzzle to try your hand at:

Miscellaneous and Personal Stuff I Must Share
1

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. :wink: 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

2

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.

3

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

4

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.

5

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. :slight_smile:

6

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.

7

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.

8

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.

9

[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? :smiley:

10

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… :slight_smile:

11

Okay, I’m using my one bump. Any other smart math people want to try this??

12

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.

13

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 = 355 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?

14

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!

15

Yeah, that’s pretty much my thoughts about it too - nice idea with the parenthetical notation. :slight_smile:

and tjdude is right too of course. :slight_smile:

16

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?

17

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 chriskerals? Can it be proven that there are none, or that there an infinite number of them?

18

If only he were that good at making football picks.

<grins, ducks, runs extremely far away> :smiley: