What number system would be the best replacement for base 10.

Doubling and Halving There is this common math problem that goes as follows: If on day 1 I get paid one penny, and each day after that my number of pennies doubles, how much money will I have at the end of thirty days? This is how many pennies you would have after each day in dozenal: 1: 1 2: 2 3: 4 4: 8 5: 14 6: 28 7: 54 8: A8 9: 194 10: 368 11: 714 12: 1,228 13: 2,454 14: 4,8A8 15: 9,594 16: 1,6B68 17: 31,B14 18: 63,A28 19: 107,854 20: 213,4A8 21: 426,994 22: 851,768 23: 1,4A3,314 24: 2,986,628 25: 5,751,054 26: B,2A2,0A8 27: 1A,584,194 28: 38,B48,368 29: 75,A94,714 30: 12B,969,228 This is how many pennies you would have after each day in octal: 1: 1 2: 2 3: 4 4: 10 5: 20 6: 40 7: 100 8: 200 9: 400 10: 1,000 11: 2,000 12: 4,000 13: 10,000 14: 20,000 15: 40,000 16: 100,000 17: 200,000 18: 400,000 19: 1,000,000 20: 2,000,000 21: 4,000,000 22: 10,000,000 23: 20,000,000 24: 40,000,000 25: 100,000,000 26: 200,000,000 27: 400,000,000 28: 1,000,000,000 29: 2,000,000,000 30: 4,000,000,000 The numbers in octal all have a 1, 2, or 4 followed by a series of zeros, while in dozenal, the answers are a seemingly random series of digits. For octal you just add a zero to the end every third day, while for dozenal, more math is needed and is impractical to do in your head or by hand. Octal makes repeatedly halving easy too. Example: 1, 0.4, 0.2, 0.1, 0.04, 0.02, 0.01, 0.004 . The simplicity of doubling and halving in octal has benefits in multiple areas including but not limited to calculating the amount of half-lives for carbon dating, and calculating the next tile number in the game 2048. Binary Conversion: Octal numbers can be converted into binary by simply converting each individual octal digit into its binary equivalence and combining them. For example, 752: 7 in octal to binary would be 111; 5 in octal to binary would be 101; 2 in binary would be 010. Put them together and you get 111101010, which is the binary equivalence of 752 octal. It"s that simple. Dozenal on the other hand, take the number 25 dozenal: You first find the biggest value that is 2^n where n is an integer, that is less than 25. That number is 14 (2^4 dozenal) Now we have 10000 (2^4 binary); you take 25 minus 14 and get 11, Now we convert 11 (dozenal) into binary 8 (2^3) is the largest 2^n number under 11, so we add 1000 to the previous 10000 to get 11000. We take 11 minus 8 to get 5 (dozenal). 5 in dozenal is 101 in binary, so we add 101 to 11000 and we get the final binary answer of 11101. We have to do much harder work to convert to binary in dozenal than in octal, even when the dozenal number is much smaller than the octal number. Computer Programming Since computers use binary, it is beneficial to use a number system in which people can easily convert to and from binary. Let"s say you have a variable that can store a positive integer of 32 bits of length; What would the largest number it can hold be? It would be 2^32 -1, which in octal is a simple easy to remember 39,999,999,999 , while the answer in dozenal would be 9BA461593 , a number almost no one is going to be able to remember. It is beneficial for programmers to know the maximum number a variable can store instead of having to look it up on a calculator. Converting base 12 "decimal", such as 0.B, to binary will always result in a number with infinite digits after the decimal point. enter 0.10000000000000 into the base n calculator cited at bottom for base 12 to base 2, and you will see the infinite 01"s at the end. Base 8 doesn"t have this problem, for example .1 octal equals .001 binary. Computers truncate infinite decimals, which creates rounding errors[1], which can lead to confusing results like .1 + .2 equaling .300000001 according to computers if dozenal were used. Source: [1] https://docs.oracle.com... (Section: Rounding Error) Base N Converter: http://korn19.ch... Calculator: http://web2.0calc.com...