Someone on 4chan once asked why the standard division algorithm starts at the left (most significant digit) of the number whereas the algorithms for addition, subtraction, and multiplication start at the right (least significant digit). Thinking about it, you can apply more or less the standard long division algorithm starting from the right so long as the last digit of the divisor is relatively prime with 10 and there is no remainder:

5634
-----
7 |39438
28
--
41
21
--
92
42
--
35
35
--
0

If there is a remainder, you could continue the division, like so:

...142857143
--------------
...00000000001|7
21
--
...98
28
--
...97
7
--
...99
49
--
...95
35
--
...96
56
--
...94
14
--
...98
28
--
...97
7
--
...99

I don't know if this is how they came up with the idea originally, but this seems to be motivating constructing p-adics.

Apparently with p-adics, they deal with divisors which have the base as a factor by allowing a finite number of digits after the radix point. For example, with 5 as a base, 1/10 would be 0.1, and 1/20 would be 0.1/2 = ...222.3. So basically the opposite of real numbers. I wonder what goes wrong with composite bases.

Thinking about 10-adics again. If you wanted to divide by something that's a multiple of 2, you would first remove all the factors of 2 from the denominator, then multiply the numerator by 0.5 a corresponding number of times. Dividing an infinite number by 2 seems possible, but I don't know if this would terminate. There could be numbers with infinitely many factors of 2 in them.

Let me define things neatly so other people can ponder it.

Given a string of base-10 digits that extends infinitely to the left, define its quotient when divided by 2 as the output of the following algorithm:

1. Start at the ones place (rightmost digit).
2. If the digit to the left of the current digit is even, divide the current digit by 2, rounding down, and let that be the corresponding digit of the result.
3. If the digit to the left of the current digit is odd, add 10 to the current digit, divide the result by 2, rounding down, and let that be the corresponding digit of the result.
4. Repeat these steps with the next digit to the left.

Is there a number other than ...000 for which this algorithm, applied repeatedly, does not eventually result in an odd number (meaning a number having 1, 3, 5, 7, or 9 as its rightmost digit)?

P10343
Found one. [spoiler: The last n digits of the number are the last n digits of [tex: 2^{\phi(5^n)}].]