#1**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**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**+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

#5**+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

#6**+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**+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

CPhill
Nov 27, 2017