+0  
 
+1
81
7
avatar

What are the divisibility rules from 1 to 13.

Guest Nov 26, 2017
Sort: 

7+0 Answers

 #1
avatar+1493 
0
Divisibility Test Number Process
1 Every integer is divisible by 1
2 Check to see if the last digit is divisible by 2
3 Recursively check if the sum of digits is divisible by 3
4 Check to see if the last two digits are divisible by 4
5 Check to see if the last digit is divisible by 5
6 Check divisibility rules for 2 & 3
7 Recursively check if subtracting twice the sum of the last digit from the rest of the number is divisible by 7
8 Check if the last 3 digits are divisible by 8
9 Recursively check if the sum of the digits is divisible by 9
10 Check if the last digit is a 0
11 Recursively check if subtracting the final digit from the rest is divisible by 11
12 Check divisibility for 3 & 4
13 Recursively add 4 times the final digit to the rest
TheXSquaredFactor  Nov 26, 2017
 #2
avatar
0

I understand the divisibility rules except 7, 11, and 13. What do you mean by the "rest?" Can you kindly explain further? 

Guest Nov 26, 2017
 #3
avatar+1493 
+1

Of course! I will gladly delve deeper into this subject.

 

The explanation I provided is probably unsatisfactory and shoddy anyway. I believe that these rules are best demonstrated by example.

 

1. Divisibility by 7

 

Is \(205226\) divisible by 7? Well, let's use the process!

 

\(\textcolor{blue}{20522}\textcolor{red}{6}\)  
1. \(\textcolor{blue}{20522}-2*\textcolor{red}{6}=\textcolor{blue}{2051}\textcolor{red}{0}\) I have no clue if this is indeed divisible by 7, so do this process again and again (hence recursion)

2. 

\(\textcolor{blue}{2051}-2*\textcolor{red}{0}=\textcolor{blue}{205}\textcolor{red}{1}\)

I still cannot tell, so I will do this again.

3.

\(\textcolor{blue}{205}-2*\textcolor{red}{1}=\textcolor{blue}{20}\textcolor{red}{3}\)

I still cannot tell.

4.

\(\textcolor{blue}{20}-2*\textcolor{red}{3}=\textcolor{blue}{1}\textcolor{red}{4}\)

I know that 14 is divisible by 7, so the original number is, too.

 

How about \(22604\)? Well, let's check it!

 

\(\textcolor{blue}{2260}\textcolor{red}{4}\)  

1.

\(\textcolor{blue}{2260}-2*\textcolor{red}{4}=\textcolor{blue}{225}\textcolor{red}{2}\)

Yet again, I cannot make a judgment.

2.

\(\textcolor{blue}{225}-2*\textcolor{red}{2}=\textcolor{blue}{22}\textcolor{red}{1}\)

Of course, we must keep going.

3.

\(\textcolor{blue}{22}-2*\textcolor{red}{1}=20\)

I know that this number is not divisible by 7, so the original number is not either.
   

 

2. Divisibility by 11

 

Let's check if \(43923\) is divisible.

 

\(\textcolor{blue}{4392}\textcolor{red}{3}\)  

1.

\(\textcolor{blue}{4392}-\textcolor{red}{3}=\textcolor{blue}{438}\textcolor{red}{9}\)

Let's do it again!

2.

\(\textcolor{blue}{438}-\textcolor{red}{9}=\textcolor{blue}{42}\textcolor{red}{9}\)

One more time!
\(\textcolor{blue}{42}-\textcolor{red}{9}=\textcolor{blue}{3}\textcolor{red}{3}\) I know that 33 is divisible by 11, so the original number is, too.
   

 

How about \(123567\)?

 

\(\textcolor{blue}{12356}\textcolor{red}{7}\)

 

1.

\(\textcolor{blue}{12356}-\textcolor{red}{7}=\textcolor{blue}{1234}\textcolor{red}{9}\)

This requires perserverance. Keep going!

2.

\(\textcolor{blue}{1234}-\textcolor{red}{9}=\textcolor{blue}{122}\textcolor{red}{5}\)

 

3.

\(\textcolor{blue}{122}-\textcolor{red}{5}=\textcolor{blue}{11}\textcolor{red}{7}\)

I know that \(11*11=121\), so 117 is not divisible.
   

 

3. Divisibility by 13

 

Is \(19704\) divisible? Let's find out!

 

\(\textcolor{blue}{1970}\textcolor{red}{4}\)  

1.

\(\textcolor{blue}{1970}+4*\textcolor{red}{4}=\textcolor{blue}{198}\textcolor{red}{6}\)

 

2.

\(\textcolor{blue}{198}+4*\textcolor{red}{6}=\textcolor{blue}{22}\textcolor{red}{2}\)

 
\(=\textcolor{blue}{22}+4*\textcolor{red}{2}=\textcolor{blue}{3}\textcolor{red}{0}\) \(13*3=36\), so 30 is not divisible by 13 and nor is the given number.
   

 

Is \(9321 \) able to be divised?
 

\(\textcolor{blue}{932}\textcolor{red}{1}\)  

1.

\(\textcolor{blue}{932}+4*\textcolor{red}{1}=\textcolor{blue}{93}\textcolor{red}{6}\)

 

2.

\(\textcolor{blue}{93}+4*\textcolor{red}{6}=\textcolor{blue}{11}\textcolor{red}{7}\)

 

3.

\(\textcolor{blue}{11}+4*\textcolor{red}{7}=\textcolor{blue}{3}\textcolor{red}{9}\)

39 is divisible by 13, so the original number is as well.
   
TheXSquaredFactor  Nov 26, 2017
 #4
avatar
+1

Thank you, once again! I can't believe you spent an hour doing this for me. 

Guest Nov 26, 2017
 #5
avatar+1493 
+1

I am glad you understand now! One unique thing about divisibility tests is that there are absolutely no exceptions to any of the rules listed. 

TheXSquaredFactor  Nov 26, 2017
edited by TheXSquaredFactor  Nov 26, 2017
 #6
avatar+6900 
+1

Easier divisibility rule for 11:

if absolute value of ((1st digit + 3rd digit + 5th digit + ....) - (2nd digit + 4th digit + 6th digit + ...)) = 0 or anything divisible by 11,

then the number is divisible by 11.

 

Example:

Is 43956 divisible by 11?

(4 + 9 + 6) - (3 + 5) = 19 - 8 = 11 <--- divisible by 11.

So 43956 is divisible by 11.

 

Is 10293583762 divisible by 11?

(1 + 2 + 3 + 8 + 7 + 2) - (0 + 9 + 5 + 3 + 6) = 23 - 23 = 0 

So 10293583762 is divisible by 11.

MaxWong  Nov 27, 2017
 #7
avatar+79894 
+1

Here's an easy rule to decide divsibility by 11

 

Alternate signs  on the digits and add.....if the result is 0  or 11, the number is divisible by 11

 

Example  43956  =   +4 - 3 + 9 - 5 + 6  = 2 + 9   = 11

 

Note .....43956  =  11 * 3996

 

Another example    1331  =  +1 - 3 + 3  - 1   =   0

 

And   1331  =   11^3

 

 

cool cool cool 

CPhill  Nov 27, 2017

15 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details