ALIEN Counting System

Question

When humans finally came across some aliens, living below the Martian crust, the strangest point was that the myth of symmetry was broken. Humans had assumed that all intelligent life forms would be symmetric is some way or the other, but our biologists report that these Martians had 31 teeth (all on the left side of the face), and 3 eyes (two in the right of the face and one in the back), and 5 fingers (in total, including 2 thumbs) and various other unsymmetricalities (Is this a real word ?).

Now, our mathematicians report that the Martian Counting System is weird too.
12 is 300
345 is 12601
67890 is 4111410
123 is 3611
456 is 16200
789 is 30501
1234 is 46410
567 is 21511

Here, the left is our system, the right is Martian System. (A few folks have come to the same conclusion, but I want to make it explicit now)

Our mathematicians want to know from you, how would 8900 be represented in that Martian System ?

EDIT:

Though 3 folks have got the answer correctly, with the numerical clues given, some minor points have not been covered, hence I will be adding the Specific Expected Answer.

Answer 1

426100

It works like binary, except: after the rightmost 3 columns (least significant digits) being 2^0 (1), 2^1 (2), 2^2 (4) (from the right to left), then each column further left is worth 3 additional powers of 2, rather than just being worth one additional power of 2.
So in binary, the 4th column from the right is the 8s column, but here the 4th column from the right is the 32s column because you jump 3 exponents to 2^5 (32). Then the 5th is 2^8 (256), the 6th is 2^11 (2048), and the 7th is 2^14 (16384).

And instead of being strictly limited to a max digit of n-1 like in a base system, you just fill up the columns as much as possible from most significant to least significant digit.

So even though the example showed a 6 as the highest possible digit, there could be a 7 in some numbers in this system, anywhere except in the final 2 digits. (e.g. take our number 31, it would need 7 in the 4 column, then 1 in the 2 column, and 1 in the 1 column.

Update: To address the comment by JLee, yes, you still actually have a unique mapping of Earth numbers to Mars numbers, assuming Martians intuitively follow digit restrictions that are natural to their number system, which they would.

It would just be obvious to them that you can only put a 1 in the ones and twos columns, but then up to a 7 in the higher columns. Note that the columns go on forever to the left, always being worth a factor of 8 more than the previous, so you can always get enough digits to get high enough to represent all numbers. Just like in our system every column is worth an additional factor of 10, and you keep going left to get numbers as big as you need.

Here is a simpler way of me summing up how the number system works.

They use a hybrid of binary and base-8. The rightmost 2 (least significant) digits are in binary, and after counting to 4, their number columns are in base-8. Basically if you chop off the right 2 digits, you have an octal representation of counting "groups of 4", and then the right 2 digits tell you whether to add 0,1,2, or 3 to that.

You could imagine such a system evolving from a Roman-Numerals like system where they used sticks to count as high as 3, but above that they used a letter to represent 4 sticks, and then another letter to represent 8 of the first letter (which represents 4 sticks), etc.

The way I did it:

After being a bit distracted by the fact that the Earth numbers were made from breaking up 1234567890, I focused on the smallest number which also happened to be simpler, with 2 zeros and just looking at 3 and 12. Well 3x4=12, is it possible that the third column could represent 4? Hey, that's like binary, but you'd never put a 3 in binary. Let's see if it's an irregular number system that still works on the "columns" idea.

I scanned the whole list and noticed that if the Martian number ended in 1 it was odd, else it was even. This is also like binary, and implies that the right column is the ones column. That means the 2nd column is probably a twos column, but I should check that it's not weird, like threes. It can't logically be anything else, since we already are thinking the 3rd column is fours (unless we are way off base).

So then I moved up to the next smallest number: 123. I played with it, subtracting the amounts if I was right about the ones, twos, and fours columns, and happily found that the remainder was divisible by 3, and the result was 32, so this must be the thirty-twos column, which is still a nice number in binary-land.

I then went on to 345 and 456 and confirmed that if the fifth column was 256, then both of these numbers worked out, and saw the pattern and then attacked the largest number to be sure it was always jumping by 2^3 per column.

Answer 2

8900 is the martian number

426100

Because

Given a decimal number X, we can represent it in quotient form X=4*Q+R. The first part of the martian number (all but the last two numbers) is Q octal, and the last two numbers are R in binary.

Example

67890
67890 % 4 = 2 = 10 (binary)
(67890-2)/4 = 16792 = 41114(octal)
so 67890 = 4111410 (martian)

Answer 3

I think that:

8900 = 426100

The Martian system uses these values for each digit:

Digit 1: $2^0$ = 1
Digit 2: $2^1$ = 2
Digit 3: $2^2$ = 4
Digit 4: $2^5$ = 32
Digit 5: $2^8$ = 256
Digit 6: $2^{11}$ = 2048
Digit 7: $2^{14}$ = 16384

Digits 1 and 2 can only be 0 or 1. All other digits can be 0-7.

BTW Starman beat me to this, but I just wanted to post it anyway.

How I arrived at this conclusion:

1. I first noticed that the end digit corresponded to whether the number was even or odd. So it acted like a binary 1's digit.
2. Next, I noticed that the next to last digit was always 0 or 1 as well. Testing indicated that it acted like the second binary digit, the 2's digit.
3. Then I used 12 = 300 to conclude that the third digit had value 4.
4. Then 123 = 3611 showed me that the fourth digit was 32.
5. Then 345 = 12601 made the fifth digit be 256.
6. Finally, 67890 = 4111410 had two digits to solve. Removing the known digits: 67584 = 4100000. Knowing that these were all powers of two, the 4 digit probably represented 65536 (4 * 16384), leaving the 1 digit to represent 2048.

Why do Martians count like this?

I'm not quite sure but maybe it's because they count using their body parts like this:

Digit 1 = Thumb 1
Digit 2 = Thumb 2
Digit 3 = Fingers 1-3
Digit 4 = Eyes 1-3
Digit 5 = Teeth 1-3 (are these teeth retractable?)
Digit 6 = Teeth 4-6
Digit 7 = Teeth 7-9
...
Digit 14 = Teeth 28-30
Digit 15 = Tooth 31

I don't know why they don't count on all their teeth all at once, but maybe it's because only digits 0-7 exist in their writing.

Answer 4

Now, Starman & Shawn Holzworth & JS1 have given correct answers, and I have accepted the answer by Starman, which was the fastest or most complete.

Here I want to Explain the Clues:

- All Martian numbers end with 00 or 01 or 10 or 11, and each ending is listed twice, hinting at some binary system in those 2 Digits.
- The remaining numbers found are 0,1,2,3,4,5,6 (having a maximum of 6, but not containing 7,8,9 and not containing letters) so that sets the lower limits to the base.
- Counting is usually (Primitively) done with fingers, and all the points about eyes and teeth are for Diverting the attention from the fingers (& to show that symmetry is not there).
- If they have 5 fingers (on 2 hands, due to 2 thumbs), then 2+3=5 may be a way the fingers are Distributed.
- If 2 fingers are used for 2 bits to represent the 2 Digits, then the remaining fingers on the other hand maybe used to represent the next Digit(s). - Octal is a good candidate for 3 fingers.
- Couple all this hand-waving (finger-waving) with the logical Explanations in the 3 other answers to figure out the Complete Alien Counting System.