This isn't a math problem exactly but I've been trying to figure it out for several months now, and no one's been able to help me. And I dont know where else to post this
So I made a base-13 number system but zero only appears alone. For example:
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, 11, 12 ...
A normal base-13 would look like this:
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, 10, 11, 12 ...
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I am trying to find a formula that converts a decimal number system (like ours) into this one. I have found one, kinda, but it only goes up to this number system's 3,468. After that, I dont know how to convert decimal numbers into this system.
This is the fomula I got for converting decimal to this system:
(x = first digit, y = second, z = third)↓
2-Digits: (y + 12) + z
3-Digits: (((x)12)(y + 12)) + z
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And on top of that, unfortunately, I dont even know how to convert numbers from this to decimal systems...
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So, I would really appreciate it if anyone could either find a formula to count past "3,468" in this system, if someone could find a way to convert this system into the decimal one, or if you can give me tips on how to do this.
Im really sorry for not giving enough information, but it's all I have as well.
Comments (2)
They are some letters that you can figure out easily if you are smart
I'll give you an example. If I can write one of your numbers with digits xyzw where X,Y,Z, and W are the base ten notation of x,y,z, and w respectively then we can write xyzw in base ten with the following formula:
X*(13^3)+Y*(13^2)+Z*(13)+W
This will let you convert numbers as high as BBBB (or 28560 in base ten) to base ten. I'll stop here and let others think about how to go backwards or extend this to even larger numbers.