I don't know if this has been posted already, but anwho... If you pick any positive integer greater than 9 and subtract the sum of its digits from that number, you'll end up with a multiple of 9. How do I know it's a multiple of 9? Curiously enough, the sum of the digits of a positive integer that is a multiple of 9 is a multiple of 9. Do this enough times and you'll end up with 9. I tried this with other number systems and it seems to hold. This is the most interesting play of numbers if found to date.
Let a_n...a_1a_0 = a_0 * 10^0 + a_1 * 10^1 + ... + a_n * 10^n be a number (in base 10). Since 10 == 1 (mod 9), we have that 10^n == 1 (mod 9). Thus a_0 * 10^0 + a_1 * 10^1 + ... + a_n * 10^n == a_0 + a_1 + ... + a_n (mod 9). Subtracting the sum of the digits, a_0 + a_1 + ... + a_n, from both sides gives the desired result. Suppose a_n...a_1a_0 = 9k for some integer k. Then a_0 * 10^0 + a_1 * 10^1 + ... + a_n * 10^n = 9k. Taking both sides modulo 9, a_0 + a_1 + ... + a_n == 0 (mod 9), as required. That's enough mathematics for today ;)
If you are not familiar with congruences and modular arithmetic, consider this... Let the number N = w + 10x + 100y + 1000z + ... (example : 574 = 4 + 7*10 + 5*100 ) Its digits are w, x, y, z, ... So the sum of the digits is w + x + y + z + ... = S, say. So, N - S = D = (w-w) + (10x-x) + (100y-y) + (1000z -z) + ... = 9x + 99y + 999z + ... = 9 (x + 11y + 111z + ...) So, N - S is a multiple of 9. Now for the second part... Consider again, N = w + 10x + 100y + 1000z + ... We want to prove that, if the sum of its digits, S is divisible by 9, then so is the number, N, and conversely. If S is divisible by 9, then S = 9p. Now to this, add the number D, calculated above. S + D = 9p + D. But we saw previously that D itself is a multiple of 9, so D = 9q. Hence, S + D = 9p + 9q = 9(p+q) = 9r, say. But then, S + D is nothing but N. So N = 9r, which is what we wnted to prove. The converse is proved by starting with N = 9r and subtracting D to yield S = 9(r-q) = 9p, say. QED
Nice proof, won't the same thing work for any base a with a-1 which is always an additive generator of the residue system? The sum of digits of any number divisible by a-1 in base a is also divisible by a-1. ???? I'm too lazy to try to crank out the proof or even find a counter to it. My question is did you gentlemen ever see the pattern of first appearance of multiples of nine when you sum the digits of multiples of nine in order? It is quite interesting. http://agapeflight.net/Summing Digits.xls Anyone know of any research related to that pattern? It appears that the distribution of digit sums would be roughly lognormal after some first appearance. I did find some stuff on research into this through the wiki on Digit Sums. Thanks.
I agree it’s very interesting. If you’d like to see some more information on this check out a previously thread called “Dissected Number Law”. Numbers 9 and 11 are key in this sort of arithmetic phenomena. https://www.physicsforums.com/showthread.php?t=248279
I'm researching that pattern - but the creator of the above link (you?) needs to take the sum of digits to completion ie Every total should have been reduced to one root digit 9 e.g 18 then add 1+ 8 = 9 27 then add 2 + 7 = 9 Anyway it is based on mod 9 This link may be helpful: http://www.applet-magic.com/Digitsum00.htm
It is called 'casting out nines', and it has been used for centuries by accountants, mathemeticians, and scribes as a simple 'sanity-check' on results. Before pocket calculators were readily available, and long- column addition and subtraction were routinely done by hand, this method was used to spot check for mistakes. Ancient scribes copying manuscripts by hand would likewise assign numeric values to specific letters, add them, and do spot-checks to keep hand-copied manuscripts free of errors as well. The earliest use of the technique I have found referenced was by Hippolytos of Rome, a bishop from the early 3rd century AD.