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.
Sunday, June 19, 2016
elementary number theory - Base system and divisibility
Subscribe to:
Post Comments (Atom)
analysis - Injection, making bijection
I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...
-
So if I have a matrix and I put it into RREF and keep track of the row operations, I can then write it as a product of elementary matrices. ...
-
I need to give an explicit bijection between $(0, 1]$ and $[0,1]$ and I'm wondering if my bijection/proof is correct. Using the hint tha...
-
Recently I took a test where I was given these two limits to evaluate: $\lim_\limits{h \to 0}\frac{\sin(x+h)-\sin{(x)}}{h}$ and $\lim_\limi...
No comments:
Post a Comment