kb/data/en.wikipedia.org/wiki/Divisibility_rule-7.md

---
title: "Divisibility rule"
chunk: 8/8
source: "https://en.wikipedia.org/wiki/Divisibility_rule"
category: "reference"
tags: "science, encyclopedia"
date_saved: "2026-05-05T08:13:56.367542+00:00"
instance: "kb-cron"
---

=== Proof using basic algebra ===
Many of the simpler rules can be produced using only algebraic manipulation, creating binomials and rearranging them.  By writing a number as the sum of each digit times a power of 10 each digit's power can be manipulated individually.

==== Case where all digits are summed ====
This method works for divisors that are factors of 10 − 1 = 9.
Using 3 as an example, 3 divides 9 = 10 − 1. That means 
  
    
      
        10
        ≡
        1
        
          
          (
          mod
          
          3
          )
        
      
    
    {\displaystyle 10\equiv 1{\pmod {3}}}
  
 (see modular arithmetic). The same for all the higher powers of 10: 
  
    
      
        
          10
          
            n
          
        
        ≡
        
          1
          
            n
          
        
        ≡
        1
        
          
          (
          mod
          
          3
          )
        
        .
      
    
    {\displaystyle 10^{n}\equiv 1^{n}\equiv 1{\pmod {3}}.}
  
 They are all congruent to 1 modulo 3. Since two things that are congruent  modulo 3 are either both divisible by 3 or both not, we can interchange values that are congruent modulo 3. So, in a number such as the following, we can replace all the powers of 10 by 1:

  
    
      
        100
        ⋅
        a
        +
        10
        ⋅
        b
        +
        1
        ⋅
        c
        ≡
        (
        1
        )
        a
        +
        (
        1
        )
        b
        +
        (
        1
        )
        c
        
          
          (
          mod
          
          3
          )
        
        ,
      
    
    {\displaystyle 100\cdot a+10\cdot b+1\cdot c\equiv (1)a+(1)b+(1)c{\pmod {3}},}
  

which is exactly the sum of the digits.

==== Case where the alternating sum of digits is used ====
This method works for divisors that are factors of 10 + 1 = 11.
Using 11 as an example, 11 divides 11 = 10 + 1. That means 
  
    
      
        10
        ≡
        −
        1
        
          
          (
          mod
          
          11
          )
        
      
    
    {\displaystyle 10\equiv -1{\pmod {11}}}
  
. For the higher powers of 10, they are congruent to 1 for even powers and congruent to −1 for odd powers:

  
    
      
        
          10
          
            n
          
        
        ≡
        (
        −
        1
        
          )
          
            n
          
        
        ≡
        
          
            {
            
              
                
                  1
                
                
                  
                    if 
                  
                  n
                  
                     is even
                  
                  ,
                
              
              
                
                  −
                  1
                
                
                  
                    if 
                  
                  n
                  
                     is odd
                  
                
              
            
            
          
        
        
          
          (
          mod
          
          11
          )
        
        .
      
    
    {\displaystyle 10^{n}\equiv (-1)^{n}\equiv {\begin{cases}1&{\text{if }}n{\text{ is even}},\\-1&{\text{if }}n{\text{ is odd}}\end{cases}}{\pmod {11}}.}
  

Like the previous case, we can substitute powers of 10 with congruent values:

  
    
      
        1000
        ⋅
        a
        +
        100
        ⋅
        b
        +
        10
        ⋅
        c
        +
        1
        ⋅
        d
        ≡
        (
        −
        1
        )
        a
        +
        (
        1
        )
        b
        +
        (
        −
        1
        )
        c
        +
        (
        1
        )
        d
        
          
          (
          mod
          
          11
          )
        
        ,
      
    
    {\displaystyle 1000\cdot a+100\cdot b+10\cdot c+1\cdot d\equiv (-1)a+(1)b+(-1)c+(1)d{\pmod {11}},}
  

which is also the difference between the sum of digits at odd positions and the sum of digits at even positions.

==== Case where only the last digit(s) matter ====
This applies to divisors that are a factor of a power of 10. This is because sufficiently high powers of the base are multiples of the divisor, and can be eliminated.
For example, in base 10, the factors of 101 include 2, 5, and 10. Therefore, divisibility by 2, 5, and 10 only depend on whether the last 1 digit is divisible by those divisors. The factors of 102 include 4 and 25, and divisibility by those only depend on the last 2 digits.

==== Case where only the last digit(s) are removed ====
Most numbers do not divide 9 or 10 evenly, but do divide a higher power of 10n or 10n − 1. In this case the number is still written in powers of 10, but not fully expanded.
For example, 7 does not divide 9 or 10, but does divide 98, which is close to 100.  Thus, proceed from

  
    
      
        100
        ⋅
        a
        +
        b
        ,
      
    
    {\displaystyle 100\cdot a+b,}
  

where in this case a is any integer, and b can range from 0 to 99.  Next,

  
    
      
        (
        98
        +
        2
        )
        ⋅
        a
        +
        b
        ,
      
    
    {\displaystyle (98+2)\cdot a+b,}
  

and again expanding

  
    
      
        98
        ⋅
        a
        +
        2
        ⋅
        a
        +
        b
        ,
      
    
    {\displaystyle 98\cdot a+2\cdot a+b,}
  

and after eliminating the known multiple of 7, the result is

  
    
      
        2
        ⋅
        a
        +
        b
        ,
      
    
    {\displaystyle 2\cdot a+b,}
  

which is the rule "double the number formed by all but the last two digits, then add the last two digits".

==== Case where the last digit(s) is multiplied by a factor ====
The representation of the number may also be multiplied by any number relatively prime to the divisor without changing its divisibility.  After observing that 7 divides 21, we can perform the following:

  
    
      
        10
        ⋅
        a
        +
        b
        ,
      
    
    {\displaystyle 10\cdot a+b,}
  

after multiplying by 2, this becomes

  
    
      
        20
        ⋅
        a
        +
        2
        ⋅
        b
        ,
      
    
    {\displaystyle 20\cdot a+2\cdot b,}
  

and then

  
    
      
        (
        21
        −
        1
        )
        ⋅
        a
        +
        2
        ⋅
        b
        .
      
    
    {\displaystyle (21-1)\cdot a+2\cdot b.}
  

Eliminating the 21 gives

  
    
      
        −
        1
        ⋅
        a
        +
        2
        ⋅
        b
        ,
      
    
    {\displaystyle -1\cdot a+2\cdot b,}
  

and multiplying by −1 gives

  
    
      
        a
        −
        2
        ⋅
        b
        .
      
    
    {\displaystyle a-2\cdot b.}
  

Either of the last two rules may be used, depending on which is easier to perform. They correspond to the rule "subtract twice the last digit from the rest".

=== Proof using modular arithmetic ===
This section will illustrate the basic method; all the rules can be derived following the same procedure. The following requires a basic grounding in modular arithmetic; for divisibility other than by 2's and 5's the proofs rest on the basic fact that 10 mod m is invertible if 10 and m are relatively prime.

==== For 2n or 5n ====
Only the last n digits need to be checked.

  
    
      
        
          10
          
            n
          
        
        =
        
          2
          
            n
          
        
        ⋅
        
          5
          
            n
          
        
        ≡
        0
        
          
          (
          mod
          
          
            2
            
              n
            
          
          
             or 
          
          
            5
            
              n
            
          
          )
        
        .
      
    
    {\displaystyle 10^{n}=2^{n}\cdot 5^{n}\equiv 0{\pmod {2^{n}{\text{ or }}5^{n}}}.}
  

Representing x as 
  
    
      
        
          10
          
            n
          
        
        ⋅
        y
        +
        z
        ,
      
    
    {\displaystyle 10^{n}\cdot y+z,}
  

  
    
      
        x
        =
        
          10
          
            n
          
        
        ⋅
        y
        +
        z
        ≡
        z
        
          
          (
          mod
          
          
            2
            
              n
            
          
          
             or 
          
          
            5
            
              n
            
          
          )
        
        ,
      
    
    {\displaystyle x=10^{n}\cdot y+z\equiv z{\pmod {2^{n}{\text{ or }}5^{n}}},}
  

and the divisibility of x is the same as that of z.

==== For 7 ====
Since 10 × 5 ≡ 10 × (−2) ≡ 1 (mod 7), we can do the following:
Representing x as 
  
    
      
        10
        ⋅
        y
        +
        z
        ,
      
    
    {\displaystyle 10\cdot y+z,}
  

  
    
      
        −
        2
        x
        ≡
        y
        −
        2
        z
        
          
          (
          mod
          
          7
          )
        
        ,
      
    
    {\displaystyle -2x\equiv y-2z{\pmod {7}},}
  

so x is divisible by 7 if and only if y − 2z is divisible by 7.

== See also ==
Division by zero
Parity (mathematics)

== References ==

== Sources ==
Apostol, Tom M. (1976). Introduction to analytic number theory. Undergraduate Texts in Mathematics. Vol. 1. Springer-Verlag. ISBN 978-0-387-90163-3.
Kisačanin, Branislav (1998). Mathematical problems and proofs: combinatorics, number theory, and geometry. Plenum Press. ISBN 978-0-306-45967-2.
Richmond, Bettina; Richmond, Thomas (2009). A Discrete Transition to Advanced Mathematics. Pure and Applied Undergraduate Texts. Vol. 3. American Mathematical Soc. ISBN 978-0-8218-4789-3.

== External links ==

Divisibility Criteria at cut-the-knot
Stupid Divisibility Tricks Divisibility rules for 2–100.