. This is something i've wondered about for a long time but never had the chance to ask an expert, or perhaps someone would direct a newbie otherwise. It all seems self evident but i feel there is also something deeper at work here: Take any number and add all the digits together, if the sum is more than one digit then again add all the digits together, repeat until the sum is one digit. Repeat for a second number, then in the same way add together the sum of the digits for both numbers: example: ; 3968, 3+9+6+8 = 26, 2+6 = 8 ; 473, 4+7+3 = 14, 1+4 = 5 ; 8+5 = 13, 1+3 = 4 add together the two original numbers and to the result sum the digits as above: ; 3968+473 = 4441, 4+4+4+1 = 13, 1+3 = 4 The answer is 4 for both sumations. This seems to will work for any number and any sum of numbers, although i dont how how to proove the summations will always be the same whatever way they are applied. Therefore if any number can be reduced to a single digit in this way, all numbers must belong to one of a number of sets that corresponds to the available digits, in this case in the range of 1-9. My question is do these sets of numbers share any other interesting properties other than the fact they correspond to the summation of their digits, and how does the set of all possible numbers distribute itself amongst the sets of reduced numbers, finally would this work for counting systems that use an base such as binary and hex or whatever. OM Thank you so much.
Here's some food for thought. Let's call this "digit sum" thing that you have D. i.e D(3968) = 8 , D(13)=4 etc. Then what you are trying to prove is whether D(x+y) = D ( D(x) + D(y) ) First define the "modulo function" . via wikipedia So 5 modulo 4 = 1. as 1 is the remainder left after trying to divide 5 by 4. 22 modulo 9 = 4, as 4 is the remainder left when trying to divide 22 by 9. etc. Can you see why D(n) = n modulo 9 when n is not a multiple of 9, and D(n) = 9 if n IS a multiple of 9? After you've sussed this, you should be able to prove D(x+y) = D( D(x) + D(y) )
A given written-out positive number, A, is a representation in a certain base, conventionally \(b= 10\) : \( A = a_n a_{n-1} a_{n-2} ... a_2 a_1 a_0 (\textrm{base} \, b) = \sum_{k=0}^{n} a_k b^k \) If we divide (with remainder) A by b, we get two numbers which can be combined to form A. \( A = a_0 + b \left\lfloor \frac{A}{b} \right\rfloor = a_0 + b \frac{A - a_0 }{b} = a_0 + b \sum_{k=0}^{n-1} a_{k+1} b^k \) But since \(b = 1 + ( b - 1 ) \), we have \( A = a_0 + \sum_{k=0}^{n-1} a_{k+1} b^k + (b - 1) \sum_{k=0}^{n-1} a_{k+1} b^k \) Thus, modulo (b-1) \( A \equiv a_0 + \sum_{k=0}^{n-1} a_{k+1} b^k \equiv a_0 + a_1 + \sum_{k=0}^{n-2} a_{k+2} b^k \equiv a_0 + a_1 + a_2 + \sum_{k=0}^{n-3} a_{k+3} b^k \equiv ... \equiv \sum_{k=0}^n a_{k+1} ( \textrm{mod} ( b - 1 ) ) \) Or for conventional base 10 numbers, the number A is equal to the sum of its digits modulo 9. If you repeat this summing of digits until you have just a single digit, that is the number's digital root. And because addition and multiplication are faithful under modulo 9, so are digital roots. \( 123 + 456 = 579 \equiv 3 \) \( 9123 + 456 = 9579 \equiv 3 \) \( 123 \times 456 = 56088 \equiv 9 \) but subtraction is tricker since \(0 \equiv 9\), \(-1 \equiv 8\), \(-2 \equiv 7\), ... \( 123 - 456 = -333 \equiv -9 \equiv 9\) \( 9123 - 456 = 8667 \equiv 9\) \( 9123 - 123 = 9000 \equiv 9\) \( 456 - 123 = 333 \equiv 9\) Division is even tricker, since \( 2035 / 5 = 407 \equiv 2\) even if \(5 \times 407 = 2035\) makes sense. http://en.wikipedia.org/wiki/Casting_out_nines http://en.wikipedia.org/wiki/Modular_arithmetic
. thanx rpenner, aleph, that's more than interesting, i had never heard of casting out nines so that helps. the wiki page redirects to Buckminster Fuller's book 'Synergetics', curious to find casting out nines in the numerology section of that book, another example of the synthesis of knowledge and intuition that the truely great mathematicians possess. see also Beoit Mandlebrot who died recently as a further exemplar of these qualities. OM
