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November 16th, 2018, 11:59 AM  #1 
Senior Member Joined: Dec 2015 From: iPhone Posts: 482 Thanks: 73  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 12:01 PM. 
November 16th, 2018, 12:11 PM  #2 
Senior Member Joined: Aug 2012 Posts: 2,260 Thanks: 687  
November 17th, 2018, 02:14 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,475 Thanks: 2039 
Consider the numerical difference between an integer and the sum of its digits.

November 17th, 2018, 02:31 AM  #4 
Senior Member Joined: Dec 2015 From: iPhone Posts: 482 Thanks: 73 
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, 05:21 AM  #5 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
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, 02:11 PM  #6 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
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 $ $3 \  \ p \ \because \ a,\ r \in \mathbb Z.$ 

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