
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
 LinkBack  Thread Tools  Display Modes 
November 16th, 2018, 12:59 PM  #1 
Senior Member Joined: Dec 2015 From: iPhone Posts: 388 Thanks: 61  Divisibility rule
If 3 divides $\displaystyle n$ for $\displaystyle n \in N$ Show that the sum of digits of $\displaystyle n$ must be divisible by 3 Last edited by idontknow; November 16th, 2018 at 01:01 PM. 
November 16th, 2018, 01:11 PM  #2 
Senior Member Joined: Aug 2012 Posts: 2,157 Thanks: 631  
November 17th, 2018, 03:14 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,303 Thanks: 1974 
Consider the numerical difference between an integer and the sum of its digits.

November 17th, 2018, 03:31 AM  #4 
Senior Member Joined: Dec 2015 From: iPhone Posts: 388 Thanks: 61 
Im not sure about this way , but the result works First express $\displaystyle n$ as powers of $\displaystyle 10$ , example: $\displaystyle 273=10^2 \cdot 2 +10^1 \cdot 7 +10^0 \cdot 3$ Let $\displaystyle x_1,x_2,...,x_k$ be the digits of $\displaystyle n$ $\displaystyle n=10^{k1}x_k + 10^{k2}x_{k1} + ...+10x_2+x_1=(10^{k1}1)x_k +x_k+(10^{k2}1)x_{k1} + x_{k1} +...+9x_2 + x_2 +x_1$ $\displaystyle 10^p 1$ is a multiple of $\displaystyle 9$, so write $\displaystyle 10^p 1=9a$ $\displaystyle n=9a_k x_k +x_k +9a_{k1}x_{k1}+x_{k1} +...+9x_2 +x_2 +x_1$ $\displaystyle n=9(a_k x_k +a_{k1}x_{k1}+...+x_2)+(x_1+x_2+...+x_k)$ $\displaystyle 39(a_k x_k +a_{k1}x_{k1}+...+x_2)$ but also $\displaystyle 3$ must divide the other part of the sum (if 3p and 3q , then 3(p+q)) so $\displaystyle 3(x_1+x_2+...+x_k)$ 
November 17th, 2018, 06:21 AM  #5 
Senior Member Joined: May 2016 From: USA Posts: 1,306 Thanks: 549 
It is hard to comment on proofs because we do not know what axioms and theorems you have available and what conventions of presentation you should follow. But the basic thought behind your proof looks sound to me. 
November 17th, 2018, 03:11 PM  #6 
Senior Member Joined: May 2016 From: USA Posts: 1,306 Thanks: 549 
I keep thinking that this argument would be much clearer with better notation. Given $m,\ n \in \mathbb Z.$ $m > 0.$ $10^{(m 1)} < n < 10^ m.$ $3\ \ n \implies \exists \ a \in \mathbb Z \text { such that } 3a = n.$ $k \in \mathbb Z \text { and } 1 \le k \le m.$ $\text {Define } d_k \text { such that } d_k \in \mathbb Z,\ 0 \le d_k \le 9, \text { and } \displaystyle \left ( \sum_{k=1}^m d_k * 10^{(k1)} \right ) = n.$ $\text {Let } p = \displaystyle \sum_{k=1}^m d_k.$ $\text {Prove } 3 \  \ p.$ $\text {Let } x_k = d_k * 10^{(k1)}  d_k \implies x_k \in \mathbb Z \ \because d_k \in \mathbb Z.$ $\text {Also } x_k + d_k = d_k * 10^{(k1)}.$ $\text { Let } q = \displaystyle \sum_{k=1}^m x_k \implies q \in \mathbb Z \ \because x_k \in \mathbb Z.$ $n = \displaystyle \left ( \sum_{k=1}^m d_k * 10^{(k1)} \right ) = \left ( \sum_{k= 1}^ m d_k + x_k \right ) = \left ( \sum_{k=1}^m d_k \right ) + \left ( \sum_{k=1}^m x_k \right ) = p + q \implies$ $p = n  q = 3a  q.$ $x_k = d_k * 10^{(k1)}  d_k = d_k(10^{(k1)}  1).$ We now apply the following theorem. $\alpha \in \mathbb Z \text { such that } \alpha \ge 1 \implies \exists \ \beta \in \mathbb Z \text { such that } 9\beta = 10^{(\alpha 1)}  1.$ $\therefore \exists \ y_k \in \mathbb Z \text { such that } 9y_k = 10^{(k1)}  1.$ $\therefore x_k = 9d_ky_k.$ $\therefore q = \displaystyle \left ( \sum_{k=1}^m 9d_ky_k \right ) = 9 * \left ( \sum_{k=1}^m d_ky_k \right ).$ $\text {Let } r = \displaystyle \left ( \sum_{k= 1}^m d_ky_k \right ) \in \mathbb Z \ \because \ d_k, \ y_k \in \mathbb Z.$ $\therefore q = 9r \implies p = 3a  9r = 3(a  3r) \implies $ $3 \  \ p \ \because \ a,\ r \in \mathbb Z.$ 

Tags 
divisibility, rule 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
General rule of divisibility  Dacu  Algebra  1  November 24th, 2014 09:42 PM 
Discrete Math: Proving A Divisibility Rule  soulrain  Applied Math  12  July 2nd, 2012 02:40 PM 
how do yo solve this one using chain rule with quotient rule  Peter1107  Calculus  1  September 8th, 2011 11:25 AM 
Simpsons rule,Trapezoidal rule,check please  manich44  Algebra  3  February 17th, 2010 10:23 AM 
should i apply the product rule or quotient rule?  mt055  Calculus  3  October 29th, 2009 11:58 PM 