I have seen the following one. Please give the proof of the observation. We know that, The difference between the sum of the odd numbered digits (1st, 3rd, 5th...) and the sum of the even numbered digits (2nd, 4th...) is divisible by 11. I have checked the same for other numbers in different base system. For example, if we want to know 27 is divisible by 3 or not. To check the divisibility for 3, take 1 lees than 3 (i.e., 2) and follow as shown bellow now 27 = 2 X 13 + 1 and then 13 = 2 X 6 + 1 and then 6 = 2 X 3 + 0 and then 3 = 2 X 1 + 1 and then 1 = 2 X 0 + 1 Now the remainders in base system is 27 = 11011 sum of altranative digits and their diffrence is ( 1 + 0 + 1) - (1 + 1) = 0 So, 27 is divisible by 3. What I want to say that, to check the divisibility of a number K, we will write the number in K-1 base system and then we apply the 11 divisibility rule. How this method is working.Please give me the proof. Thanks in advance.
Base system and divisibility
3
$\begingroup$
elementary-number-theory
arithmetic
-
3That's a keen observation. You might want to look at [modular arithmetic](http://en.wikipedia.org/wiki/Modular_arithmetic). Also, just as you generalized from $11$ to $b+1$ in base $b$, there are generalizations of the divisibility rules for $2$ and $5$ to the divisors of $b$ and of the divisibility rules for $3$ and $9$ to the divisors of $b-1$. – 2011-07-19