Divisibility Rules
 
David W. Hansen
 
([email protected])
 
 
                              In this paper, we derive rules which allow us to determine whether or not a given
                                          natural number N is divisible by any of the integers 1 through 25.
 
December 20, 2007
  
  
        In dividing integers, there are several interesting (and useful) rules that are used to determine if a given  
  
integer is or is not a divisor of a given number. For example, we know that
 
                                                  a) any integer whose last digit is even is divisible by 2,
                                                  b) any integer whose last digit is 0 or 5 is divisible by 5,
                                                  c) any integer whose last digit is 0 is divisible by 10,
   and                                        d) if the sum of the digits of a number is divisible by 3, then the          
                                                       number itself is divisible by 3.
        But why are these rules true? Can they be proved? And are there rules for other divisors that we are not 
    familiar with?
        To answer these questions, we must first realize that when we write down an integer using its digits, we 
   are really writing the number as a sum of powers of ten.        35,237 is just a short-hand way of writing 
                                                                      3x104 + 5x103 + 2x102 + 3x101 + 7.
        In general, any positive integer N can be written as
                                                                   N =   an10n + an-110n-1 + . . . + a1101 + a0,
   where the a’s are positive integers or zero.  Note that each term of the sum above for N, except the last 
   one, has a factor of 10 = 5 x 2, so 2, 5, and 10 are divisors of every term of N except its last term (digit) 
   a0. Now,
            a) If this last digit, a0, is even, then it is divisible by 2, and since every other term of N is divisible by 2, 
                 then N itself is divisible by 2, proving rule a) above.
             b) If this last digit a0 is 0 or 5, then it is divisible by 5, and since every other term of N is divisible by 5,
                  then N itself is divisible by 5, proving rule b) above.  
             c) Finally, if the last digit a0 of N is 0, then it vanishes and each remaining term of N is divisible by 10,
                  so N itself is divisible by 10, proving rule c) above.  
        Now, what about rule d)? It can be stated as follows:
 

                 Division Rule for 3, Part A          

 
                 If the sum of the digits of a natural number N is divisible by 3, then N itself is divisible by 3.

              _______________________________________________________________________

 
        To prove this, we let    N =  an10n + an-110n-1 + . . . + a1101 + a0 be a natural number.    Since the sum of 
   the digits of N is divisible by 3, then
 
                                                         an + an-1 + . . . + a1 + a0  =  3k for some integer k.
 
        Solving for an,  we have   an    =    3k –   an-1  – . . . – a1 –   a0.
 
   Then,                  N       =               an10n                                 +    an-110n-1  +  . . .  +  a1101   +  a0
 
                              =  (3k – an-1  – an-2 – . . . – a1 – a0)10n     +    an-110n-1  +  . . .  +   a1101  +   a0
 
         =   3k10n  –  an-110n  –  an-210n  – . . . – a110n – a010n  + an-110n-1 +  an-210n-2 + . . .  + a1101  +  a0
 
         =   3k10n – an-1(10n – 10n-1) – an-2(10n – 10n-2) – . . . – a1(10n – 101)  –  a0(10n – 1)
 
         =    3k10n – an-110n-1(10 – 1) – an-210n-2(102 – 1) – . . . – a1101(10n-1 – 1) – a0(10n – 1)
 
         =     3k10n – an-110n-1(9) – an-210n-2(99) – . . . – a1101(99 . . . 99) – a0(99 . . . 99)
                                                                                                         (n-1) 9’s                n 9’s
 
         =    3 [ k10n – an-110n-1(3) – an-210n-2(33) – . . . – a1101(33 . . . 33) – a0(33 . . . 33) ]
                                                                                                              (n-1) 3’s              n 3’s
 
        So, N has a factor of 3 and is thus divisible by 3, proving the Division Rule for 3, Part A, above.   The
   converse of this rule is given below:
 

               Division Rule for 3, Part B 
 
               If a natural number N is divisible by 3, then the sum of its digits is divisible by 3.
             ______________________________________________________________ 
 
     To prove this rule, we proceed as follows.
 
     Since N is divisible by 3,    N = an10n  + an-110n-1 + . . . + a2102 + a1101 + a0  =  3k
 
for some integer k.   Solving this for a0, we have     a0   =   3k – an10n – an-110n-1 –  . . . – a2102 –  a1101.
 
Now, the sum S of the digits of N is    S = an + an-1  +   . . .  a2 + a1 + a0,     and substituting the value of a0 into S we get        S = an + an-1  + . . . + a2 + a1 + (3k – an10n – an-110n-1 –  . . . – a2102  – a1101)
 
                           = 3k –  an(10n – 1) –  an-1(10n-1 – 1)   – . . . –   a2(102 – 1) –   a1(101 – 1)
 
                           = 3k   – an(99 . . . 99) – an-1(99 . . . 99)   – . . . –   a2(99) –   a1(9)
                                                  n 9’s                      (n-1) 9’s
       
                            = 3 [k   – an(33 . . . 33) – an-1(33 . . . 33)   – . . . –   a2(33) –   a1(3) ]
                                                       n 3’s                    (n-1) 3’s
 
     So, the sum of the digits S has a factor of 3 and is thus divisible by 3, proving the Division Rule for 3,
  Part B.   We shall combine Parts A and B of the Division Rule for 3 into the following:
 

                Division Rule for 3          

 
                If the sum of the digits of a natural number N is divisible by 3, then N itself is divisible by 3,
                and conversely.

             _______________________________________________________________________

 

 
     Closely related to the Division Rule for 3 is the Division Rule for 9 which states:
 
                Division Rule for 9
 
                If the sum of the digits of a natural number N is divisible by 9,  then N itself is divisible by 9,
                and conversely.
             ________________________________________________________________________ 
 
     This rule can be proved by simply replacing 3k by 9k in the proofs for the Division Rule for 3 above.
 
  Example.  Is 89,467,656 divisible by 3?  To see if it is, we use Division Rule 3 and find the sum of its
  digits;  8 + 9 + 4 + 6 + 7 + 6 + 5 + 6 = 51.  Now, is 51 divisible by 3?   Again, we find the sum of its digits;
  5 + 1 = 6, which is divisible by 3, so 51is divisible by 3.  And since 51 is divisible by 3,  then 89,467,656 is
  divisible by 3. However, this number is not divisible by 9 since the sum of its digits, 51, is not divisible by 9.
 
  Example.  Here is an easy proof that the 4-digit number N = abcd is divisible by 9 if the sum of its digits
  a, b, c, and d is divisible by 9.    Since N = abcd, then
 
                   N    =    1000a   +  100b  +  10c  +  d     =     999a  +     99b  +     9c   +   (a + b + c + d).
 
  Clearly, each of the first three terms in the expression on the right is divisible by 9, so if the last term
  (a+b+c+d) is divisible by 9, then all terms of N are divisible by 9, and hence N is divisible by 9.
 
     Division Rule for 4
     To find a division rule for 4, let’s write down the expression for any 6-digit number;
                         N  =     a5105 + a4104 + a3103 + a2102 +  a1101  +  a0
                              = 100,000a5 + 10,000a4 + 1,000a3 + 100a2 + 10a1 + a0
                              =   4(25,000)a5 + 4(2,500)a4 + 4(250)a3 + 4(25)a2 + (10a1 + a0)   
  As we can see, every term except the last one (10a1 + a0) has a factor of 4, so 4 is a divisor of each term
  of N except 10a1 + a0. Thus, if 4 is a divisor of 10a1 + a0, then every term of N will be divisible by 4, and 
   thus  N will be divisible by 4.  
       In general,    N = an10n + an-110n-1 + . . . + a1101 + a0 
         =   (100 . . . 000)an + (100 . . . 000)an-1 + . . . +  1000a3 + 100a2  + 10a1 + a0                
         =   4(250 . . . 00)an +  4(250 . . . 00)an-1  +  . . . +  4(250)a3 + 4(25)a2  + (10a1 + a0)
   which shows that every term of N has a factor of 4 except (10a1 + a0).  If 4 is a divisor of (10a1 + a0),
   then every term of N will be divisible by 4, and thus N will be divisible by 4. This gives us

              Division Rule for 4

 
              If the number formed by the last two digits of a natural  number N is divisible by 4,
              then N is divisible by 4.

          __________________________________________________________________

    Example.  935,277,336 is divisible by 4 because the number formed by its last two digits, 36, is

    divisible by 4.

                                                                            Division Rule for 6
        Since 6 = 2(3), any natural number N divisible by 6 must be divisible by both 2 and 3. Thus, it must be
     both an even number and one whose sum of its digits is divisible by 3.              
           Division Rule for 6
           Any number whose sum of its digits is divisible by 3  and whose last digit is even is divisible by 6.
        ____________________________________________________________________________
         
                                                                           Division Rule for 8
     To find a division rule for 8, let’s write down the expression for any 6-digit number;
                            N  =  a5105 + a4104 + a3103 + a2102 + a1101 + a0
                                 =  100,000a5 + 10,000a4 + 1,000a3 + 100a2 + 10a1 + a0
                                 =   8(12,500)a5 + 8(1,250)a4 + 8(125)a3 + (100a2 + 10a1 + a0)   
   As we can see, every term except the last three (100a2 + 10a1 + a0) has a factor of 8, so 8 is a divisor of
   each term of N except the last three, 100a­2 + 10a1 + a0. Now, if 8 is a divisor of 100a2 + 10a1 + a0, then 
   every term of N will be divisible by 8, and thus N will be divisible by 8.  
         In general,    N =  an10n + an-110n-1 + . . . + a1101 + a0 
                                   =   (2)n(5)nan + (2)n-1(5)n-1an-1 + . . . +  (2)3(5)3a3 + (2)2(5)2a2 + (2)(5)a1  + a0                
                                   =   2n(5nan) +  2n-1(5n-1an-1) + . . . +  23(53a3) + (100a2  + 10a1 + a0),
   which shows that every term of N has a factor of 8 (for n > 2) except (100a2 + 10a1 + a0). Now, if 8 is a 
   divisor of (100a2 + 10a1 + a0), then every term of N will be divisible by 8, and thus N will be divisible by 8, 
   giving us

                  Division Rule for 8

                 
                  If the number formed by the last three digits of a natural number N is divisible by 8,
                  then N is divisible by 8.

               _________________________________________________________________

  Example. 35,816 is divisible by 8 because the number formed by its last three digits, 816, is divisible by

  8.  It is not divisible by 6 because even though it is even, the sum of its digits, 3 + 5 + 8 + 1 + 6 = 23, is not

  divisible by 3.

    
     Now, what about division by 7? Can we find a rule which tells us whether or not a number is divisible by 7?  
 
                                                                       Division Rule for 7
 
        Consider 14. It is divisible by 7, but the sum of its digits (1 + 4 = 5) is not. However, 3 times its first
    digit, 3(1), plus its last digit, 4, equals 7, which is divisible by 7.  
 
                                                                 14   =  14    →    3(1) + 4 = 7
 
        Consider 21. It is divisible by 7, but the sum of its digits (2 + 1 = 3) is not. However, 3 times its first
     digit, 3(2), plus its last digit, 1, is 7, which is divisible by 7.   
 
                                                                  21   =   21    →    3(2) + 1 =  7
 
         Consider 35. It is divisible by 7, but the sum of its digits (3+5 = 8) is not. However, 3 times its first
      digit, 3(3),  plus its last digit, 5, is 14, which is divisible by 7.      
 
                                                                   35   =   35    →   3(3) + 5 =  14
 
          Consider 49. It is divisible by 7, but the sum of its digits (4+9 = 13) is not. However, 3 times its first
       digit, 3(4),  plus its last digit, 9, is 21, which is divisible by 7.  
 
                                                                   49   =   49    →    3(4) + 9 = 21
 
           Consider 119. It is divisible by 7, but the sum of its digits (1 + 1 + 9 = 11) is not. However, 3 times its
         first  two digits, 3(11), plus its last digit, 9, equals 42, which is divisible by 7.
         
                                                                 119   =   119    →    3(11) + 9 = 42
 
           Consider 553. It is divisible by 7, but the sum of its digits (5 + 5 + 3 = 13) is not. However, 3 times its
         first  two digits, 3(55), plus its last digit, 3, equals 165 + 3 = 168.
 
                                                                  553   =   553    →    3(55) + 3 = 168
           Now, is 168 divisible by 7? Let’s apply our rule.  3 times its first two digits, 3(16), plus its last digit, 8,
     equals 48 + 8 = 56, which is divisible by 7. Thus, if our rule is true, 168 is divisible by 7, and if 168 is
     divisible by 7, then so is 553. (It is, since 553/7 = 79.)   In condensed form,
 
                                                 553  =  553    →    3(55) + 3   =  168    →    3(16) + 8  =  56
 
         These examples seem to suggest that for any number N, if we
 
                        a) multiply by 3 the number formed by all of the digits of N except its last, and then
 
                        b) add the last digit of N to this result,
 
          then, if the number obtained in b) is divisible by 7, then N will be divisible by 7.
 
       Let’s try this for N  =  434  =  434  →  3(43) + 4  = 133.       Continuing in the same way, we get                
                                         133  =  133  →  3(13) + 3   =  42,
                                            42  =   42   →  3(4)  +  2   = 14,  
                                            14  =   14   →  3(1)  +  4   =    7.
 
       Now 7 is clearly divisible by 7, so N = 434 is divisible by 7. True, since 434 = 7(62).
 
          To prove this Division Rule for 7, let N be any natural number. Then,
 
                                          N  =    an10n  +  an-110n-1  +  . . . +  a1101  +  a0
 
                                               =  10(an10n-1  +  an-110n-2  +  . . . + a1)  +  a0
 
                                               =             10A     +    L1 ,
 
       where  A  is the number formed by ALL of the digits of N except its last one, and L1 is the last digit of N.
 
       For example, 27,453 = 10(2,745) + 3 = 10A + L1, where A = 2,745, the number formed by ALL the digits
    of 27,453 except its last, and L1 = 3, the last digit of 27,453.
 
                                 Now, let           N  = 10A + L1. 
 
    Then,                                             N  =   7A + 3A + L1   =  7A + (3A + L1)
 
    Looking at the right-hand side of the equation above, we see that the first term, 7A , is clearly divisible by 7,
    so if the second term,  3A + L1 , is divisible by 7, then N will be divisible by 7. This gives us the

 

             1st Division Rule for 7 (3A + L1)
 
              If N   = 10A + L1, where A is the number formed by ALL of the digits of N except its last one,
              and L1 is the last digit of N, then N is divisible by 7 if 3A + L1 is divisible by 7.
           _________________________________________________________________________
 
         Let’s use this rule to see if 2,422 is divisible by 7.
 
                                           2422  =  2422   →   3(242) + 2  =  726 + 2  =  728
 
                                              728  =    728   →     3(72) + 8  =  216 + 8  =  224
 
                                              224  =    224   →     3(22) + 4  =     66 + 4  =   70
 
      Since 70 is divisible by 7, then 2,422 is divisible by 7.
 
                                                                   A Second Division Rule for 7
 
         In a manner similar to what we’ve done above, we can rewrite N as follows.
 
                                  N  =  an10n + an-110n-1 + . . . + a1101 + a0
 
                                       =  100(an10n-2 + an-110n-3 + . . . + a2) + 10a1 + a0
 
                                       =             100A     +    L2 ,
 
    where A is the number formed by ALL of the digits of N except its last two, and L2 is  the number formed
    by the last two digits of N.
 
        For example, 27,453 = 100(274) + 53 = 100A + L2, where A = 274, the number formed by ALL the digits
    of 27,453 except its last two, and L2 = 53, the last two digits of 27,453. 
     
                        Now, let   N   =       100A    +  L2.  
 
    Then,                            N   =   98A + 2A + L2   =   7(14)A   +   (2A + L2)
 
          Looking at the far right-hand side of the equation above, we see that the first term, 7(14)A , is clearly
     divisible by 7, so if the second term,  2A + L2,  is divisible by 7,  then N will be divisible by 7.  This gives
     us the
 
               2nd Division Rule for 7 (2A + L2)
 
                If N  = 100A + L2, where A is the number formed by ALL of the digits of N except its
                last two, and L2 is the number formed by the last two digits of N, then N is divisible by 7
                if 2A + L2 is divisible by 7.
             ______________________________________________________________________
    
       Let’s try this 2nd rule on  N  =  4473  =  4473  →  2(44) + 73  =  88 + 73  =  161 
 
                                                             161  =   161   →    2(1) +  61  =  63.
 
     Now 63 is clearly divisible by 7, so N = 4473 is divisible by 7. 
  
                                                                      A Third Division Rule for 7
 
            In a manner similar to what we’ve done twice before, we can rewrite N as follows.
 
                                           N  =  an10n + an-110n-1 + . . . + a1101 + a0
 
                                                =  1000(an10n-3 + an-110n-4 + . . . + a3) + 100a2 + 10a1 + a0
 
                                                =             1000A     +    L3,
 
    where A is the number formed by ALL of the digits of N except its last three, and  L3 is the number formed
    by the last three digits of N.
 
        For example, 27,453 = 1000(27) + 453 = 1000A + L3, where A= 27, the number formed by ALL the
    digits of 27,453 except its last three, and L3 = 453, the last three digits of 27,453.
 
         Now, let           N  =  1000A + L3. 
 
    Then,                    N   =  1001A  –  A  +  L3    =   7(143)A  –   (A  –  L3).
 
         Looking at the right-hand side of the equation above, we see that the first term, 7(143) A, is clearly
     divisible by 7, so if the second term, A – L3, is divisible by 7, then N will be divisible by 7. This gives us the
  

               3rd Division Rule for 7 (A – L3)

 
                If N  = 1000A + L3, where A is the number formed by ALL of the digits of N except its last
                three, and L3 is the number formed by the last three digits of N, then N is divisible by 7
                if A – L3 is divisible by 7.

            ________________________________________________________________________

 
         Let’s try this 3rd rule on    N   =   667,184  =   667184   →   667  –  184   =  483.
 
     Using the 2nd rule, we have                    483  =         483   →    2(4)  +   83   =    91,
 
     and, by the 1st rule, we get                       91  =           91    →    3(9)  +    1    =    28. 
 
    Since 28 is clearly divisible by 7, so N  =  667,184 is divisible by 7. 
 
      Now, in the 1st Rule for division by 7; namely, 3A+L1, we must multiply A by 3, which could be a bit lengthy
   if A is a large number. Similarly, in the 2nd Rule, 2A+L1, we must multiply A by 2. However, in the 3rd Rule,
   A – L3, we need only to subtract L3 from A and do not need to multiply A by any number at all. This suggests
   that we should try to find rules for division in the form of A – (some multiple of L1, L2, or L3) to make them
   easier to apply. Thus, we shall look for rules of division in the form of  A – xL1, where x is an integer.      
 
                                                                         More Division Rules for 7
 
               Let N = 10A + L1.    Solving this for L1, we get L1 = N – 10A.                                                              (1)
 
   Now,          N  =  10A + L1  =  9A  + L1 + (A – xL1) + xL1, where x is an integer.
 
   Then,                                     N     =    9A + (x + 1)L1 + (A – xL1)
 
    Substituting for L1 from (1),   N     =     9A + (x + 1)(N – 10A) + (A – xL1)
 
                                              N – 9A    =   (x + 1)N – 10A(x + 1) +  (A – xL1)
 
                                 10A(x + 1) – 9A   =   (x + 1)N   –   N   +   (A – xL1)
 
                                          10Ax  +  A   =   xN  +  N  –  N  +  (A – xL1)
 
                      (10x + 1)A – (A – xL1)   =   xN       
 
    Now, if we choose x so that 10x + 1 and A – xL1 are both divisible by 7, then the left-hand side of the above
  equation will be divisible by 7, and thus the right-hand side, xN, will be also. Now, if x were divisible by 7,
  then10x would also be divisible by 7, and 10x+1 would not be divisible by 7. But this contradicts the fact that
  we chose x so that 10x + 1 would be divisible by 7. So x cannot be divisible by 7, and N must be divisible by
  7, since 7 is a prime. 
 
       Note:  If 10x + 1 is divisible by 7, then 10x + 1 is a multiple of 7, so 10x + 1 = 7k for some integer k.
  Solving for x, we get x = (7k – 1)/10.                                              
   
       Furthermore, once we find A – xL1, we can find other division rules for 7 by adding multiples of 7L1
   to A – xL1, because for any integer k1, k1(7L1) is divisible by 7, and if A – xL1 is divisible by 7, then
   so is  (A – xL1)  +  k1(7L1).  All this gives us
 

         General Division Rule for 7

   
         Let N = 10A + L1. Then, N is divisible by 7 if A – xL­1­ is divisible by 7, where x is an integer selected so
         that 10x + 1 is divisible by 7; that is,  x = (7k – 1)/10  for some integer k.  Furthermore, if A – xL1 is 
         divisible  by 7, then so is  A – xL1  +  k1(7L1)  for any integer k1.

      ________________________________________________________________________________ 

  
   Now, to find x, we must find a value for k which makes x = (7k – 1)/10 an integer.
 
        For k = 1, we have x  =  (7 – 1)/10  =  6/10.  For k = 2, we have x  =  (14 – 1)/10  =  13/10.
 
        For k = 3, we have x  =  (21 – 1)/10  =  2,  an integer.   So, x = 2, and we get   A – xL1 = A – 2L1,
        which gives us the
 

                    4th Division Rule for 7 (A – 2L1)

 
                    If N   = 10A + L1, where A is the number formed by ALL of  the digits of N except its last one,
                    and L1 is the last digit of N,  then N is divisible by 7 if A – 2L1 is divisible by 7.
                _________________________________________________________________________
 
     Let’s try this 4th rule on 130,564.
 
                 N  =  130,564  =  130564    →   13056 – 2(4)  =  13056 – 8   =  13048
 
                              13048  =   13048     →     1304 – 2(8)  =   1304 – 16  =  1288
 
                                1288  =     1288      →       128 – 2(8)  =     128 – 16  =    112
                 
                                   112  =      112       →         11 – 2(2)  =         11 – 4   =      7.
 
      Now, 7 is clearly divisible by 7, so N = 130,564 is divisible by 7. 
 
         We know that by using the General Rule for Division by 7, we can add multiples of  7L1 to the rule
     A – 2L1, giving us new rules for the division of 7.  Thus, by adding 7L1  to  A – 2L1,  we obtain the new
     rule,  A + 5L1.  Let’s try this new rule on 130,564.
 
                                  N  =  130,564  =  130564    →   13056 + 5(4)  =  13056 + 20  =  13076
 
                                              13076  =    13076     →     1307 + 5(6)  =    1307 + 30  =    1337
 
                                                1337  =      1337      →      133 + 5(7)  =      133 + 35   =     168
                                         
                                                  168  =        168       →        16 + 5(8)  =        16 + 40   =       56.
 
      Now, 56 is clearly divisible by 7, so N = 130,564 is divisible by 7. 
 
          To summarize all these new rules, we have
 

                    Division Rules for 7

 
                    Let N be a natural number and L1, L2, and L3 be the numbers formed by its last digit,
                    its last two digits, or its last three digits, respectively. Let A be the number formed by
                    ALL the remaining digits of N taken in order. Then, N is divisible by 7 if any of the
                    following expressions are divisible by 7:
 
                                 a)   3A + L1,    A – 2L1,    A + 5L1,         b)   2A + L2,          c)   A – L3.                 

                  _______________________________________________________________________

 
                                                                           Divisibility Rules
 

          1. Any number is divisible by 1.

 

          2. Any number whose last digit is even is divisible by 2.

 

          3. If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3.

 

          4. If the number formed by the last two digits of a number N is divisible by 4, then N is divisible by 4.

   
          5. Any number whose last digit is 0 or 5 is divisible by 5.
 
          6. Any number whose sum of its digits is divisible by 3 and whose last digit is even is divisible by 6.
 
          7.  If N = 10A + L1, where A is the number formed by ALL of the digits of N except its last one, and
               L1 is the last digit of N, then N is divisible by 7 if A – 2L1 is divisible by 7. (See Division Rules for 7
               above for more rules.)   
 
     8.   If the number formed by the last three digits of a number N is divisible by 8, then N is divisible by 8.
 
     9. If the sum of the digits of a number is divisible by 9, then the number itself is divisible by 9.
 
          10. Any number whose last digit is 0 is divisible by 10.     
       ________________________________________________________________________________                             
    
        To find division rules for integers greater than 10, let’s follow the same steps as we did in proving the
   General Division Rule for 7, but this time let’s find a general division rule for any divisor D, where D is
    prime.           
 
                                                   General Division Rule for Any Prime Divisor D
 
           Let    N  =  10A + L1.      Solving this for L1, we get L1  =  N – 10A.                                                          (2)
 
     Now,       N  =   10A + L1  =   9A  +  L1 +  (A – xL1)  +  xL1,  where x is a positive integer.
 
     Then,                                           N     =    9A  +  (x + 1)L1  +  (A – xL1).
 
      Substituting for L1 from (2),     N     =     9A  +  (x + 1)(N – 10A)  +  (A – xL1)
 
                                                  N – 9A    =   (x + 1)N  –  10A(x + 1)  +  (A – xL1)
 
                                    10A(x + 1) – 9A   =   (x + 1)N   –   N   +   (A – xL1)
 
                                                10Ax + A   =   xN  +  N  –  N  +  (A – xL1)
 
                         (10x + 1)A  –  (A – xL1)   =   xN       
 
        Now, if we choose x so that 10x + 1 and A – xL1 are both divisible by D, then the left-hand side of the
     above equation will be divisible by D, and thus the right-hand side, xN, will be also. Now, if x were divisible
     by D, then 10x would also be divisible by D,  and  10x+1  would not be divisible by D. But this contradicts
     the fact that we chose x so that 10x + 1 would be divisible by D. So x cannot be divisible by D, and N
     must be divisible by D, since D is prime.
 
         Thus, if A – xL1 is divisible by D, where x is chosen so that 10x+ 1 is divisible by D, then N will be
      divisible by D.
 
       Note:  If 10x + 1 is divisible by D, then 10x + 1 is a multiple of D, so 10x + 1 = kD for some integer
       k. Solving for x, we get   x = (Dk – 1)/10. Furthermore, once we find  A – xL1, we can find other division
       rules for D by adding multiples of  DL1  to  A – xL1,  because for any integer k1,  k1DL1 is divisible by D,
       and if A – xL1 is divisible by D,  then so is  (A – xL1)  +  k1DL1.  All this gives us
 

                 General Division Rule for Any Prime Divisor D
            
             Let N = 10A + L1. Then, N is divisible by D if A – xL­1­ is divisible by D, where x is an integer 
             selected so that 10x + 1 is divisible by D;  that is,  x  =  (Dk – 1)/10  for some  integer k. 
             Furthermore,  if  A – xL1  is divisible by D, then so is  A – xL1 +  k1DL1  for any integer k1.
            ______________________________________________________________________
 
                                                                         Division Rule for 11
 
       Let’s find a Division Rule for 11. Using the General Division Rule above, we have  D = 11, and we must
  find k so that x = (Dk – 1)/10 = (11k – 1)/10 is an integer. Starting with k = 1, we get x = (11 – 1)/10 = 1. So,
  x = 1,  A – xL1  =  A – L1,  and we have a
 
                      Division Rule for 11      If A – L1 is divisible by 11, then so is N.                      
                     ___________________________________________________

 

     Example.      41,228  =  41228    →   4122 – 8  =  4114
 
                                                4114    →    411 – 4   =     407
                  
                                                   407   →      40 – 7   =       33
                         
     Since 33 is divisible by 11, so is 41,228.     41,228 = 7( 3,748).    
 
     Here’s another way to arrange the subtractions.                
 
                                                                   41228      
                                                                   4122
                                                                        -8         
                                                                  4114    
                                                                     -4
                                                                  407
                                                                   -7
                                                                  33
 
 
                                                                        Division Rule for 12
 
         Since 12 = 4(3), any natural number N divisible by 12 must be divisible by both 4 and 3. Thus, its last
   two digits must be divisible by 4, and the sum of its digits must be divisible by 3.              
  
                           Division Rule for 12
 
                           Any number whose sum of its digits is divisible by 3 and whose last two
                           digits are divisible by 4 is divisible by 12.

                         ________________________________________________________

 

    Example.  14,686,308 is divisible by 12 because the sum of its digits, 36, is divisible by 3 and its last
     two digits, 08, is divisible by 4.  In fact, 14,686,308  ÷  12  =  1,223,859.
 
                                                                                   Division Rule for 13
 
       Using the General Division Rule above, we have D = 13, and we must find k so that x = (13k – 1)/10 is
   an integer. To find k, we construct the following table.
  
                                                          k     13k – 1    x = (13k – 1)/10
                                                   --------------------------------------------------
                                                          1           12                1.2
                                                          2           25                2.5
                                                          3           38                3.8
                                                          4           51                5.1
                                                         -    -    -    -    -    -    -    -   -           
                                                          7           90                  9         

             So, k  =  7, x  =  9,  and  A – xL1  =  A – 9L1.
 
       To get more rules, we can use the General Rule for Division and add multiples of 13L1  to  A – 9L1. 
    Thus, we can add 13L1  to  A – 9L1,  getting A + 4L1  as a new rule, or  we can add 2(13L1)  =  26L1
     to  A – 9L1  getting  A + 17L1  as another rule. This gives us
                       
                        Division Rule for 13:    If  A – 9L1,   A + 4L1,   or    A + 17L1
                        is divisible by 13, then so is N.                 

                      ________________________________________________

      
    Example.   Is 816,621 divisible by 13?    Using A – 9L1, we get   
                                                                                                                                             816621
                                                                                                                                             81662
                           816,621  =  816621   →   81662 – 9(1)  =  81653                              ­      - 9
                                                                                                                                            81653
                                                  81653   →     8165 – 9(3)  =    8138                                -27 
                                                                                                                                           8138
                                                     8138   →      813 – 9(8)   =      741                             -72 
                                                                                                                                           741
                                                       741   →         74 – 9(1)   =       65                              -9
                                                                                                                                           65                                           
                                                                               
     Since 65 is divisible by 13, so is 816,621.    (816,621 ÷ 13  =  62,817)                                                                                                                                               
 
                                                                   Division Rule for 14
 
         Since 14 = 2(7), any natural number N divisible by 14 must be divisible by both 2 and 7.  Thus, N must
    be even,  and  A – 2L1  must be divisible by 7.

                      Division Rule for 14

                      If N is even and A – 2L1 is divisible by 7, then N is divisible by 14.

                

             Example.  11,702,544 is divisible by 14 because it is even, and using  A – 2L1  or 2A + L2, we get
                                                    A – 2L1                                               2A + L2
 
                                              11,702,544                                            11,702,544
                                                           - 8                                        2A       234,050
                                              1,170,246                                         L2             + 44
                                                       - 12                                                     234,094
                                               117,012                                          2A           4,680        
                                                       - 4                                             L2             + 94
                                               11,697                                                            4,774     
                                                  - 14                                              2A                 94
                                               1,155                                               L2             + 74
                                                - 10                                                                   168
                                                105                                                2A                  2                                   
                                              - 10                                                   L2            + 68       
                                                  0,                                                                      70,
                                 which is divisible by 7.                               which is divisible by 7.
                                                        
      In fact, 11,702,544 ÷ 14  =  835,896.
 
                                                                     Division Rule for 15
 

         Since 15 = 3(5), any natural number N divisible by 15 must be divisible by both 3 and 5. Thus, the sum

    of its digits must be divisible by 3, and its last digit must be either 0 or 5.

 
                         Division Rule for 15
 
                         If the last digit of N is 0 or 5, and the sum of its digits is divisible by 3,
                         then N is divisible by 15.
                       _______________________________________________________ 
 
    Example.  5,774,655 is divisible by 15 because its last digit is 5, and the sum of its digits, 39, is divisible
    by 3.  In fact, 5,774,655 ÷ 15  =  384,977.
 
                                                                     Division Rule for 16
 
          Looking at the division rules for 4 and 8 and their proofs, we see that we can generalize these proofs to
    get division rules for higher powers of 2. Doing this, we get
 
                      Division Rule for 16
 
                      If the number formed by the last four digits of a natural number N is divisible by 16,
                      then N is divisible by 16.
                     _______________________________________________________________
 

                        Division Rule for 2n

 
                        If the number formed by the last n digits of a natural  number N is divisible by 2n,
                        then N is divisible by 2n.
                      _______________________________________________________________
 
    Example.  Is 1,205,584 divisible by 16?  This not an easy problem because by our rule, we must determine
    if the number formed by the last 4 digits, namely, 5,584, is divisible by 16.  And how do we do that?  By
    dividing 5,584 by 16.  But in using any of our other division rules, we have never had to divide. So, this rule
    is of limited value.  It saves us some work, since we need divide only the 4-digit number, 5584, by 16
    instead of dividing the original 7-digit number 1,205,584 by 16. To do this, we divide 5,584 four times by 2,
    giving us  2,792;  1,396;  698;  and  349.  So, 5,584 is divisible by 16, and thus so is 1,205,584. 
 
                                                               Division Rule for 17
 
        Let’s find a Division Rule for 17. Using the General Division Rule above, we have D = 17, and we must
      find k so that x = (Dk – 1)/10 = (17k – 1)/10 is an integer. To find k, we construct the following table.
                                                               k       17k – 1     x = (17k – 1)/10
                                                            -----------------------------------------------
                                                                1           17                1.7
                                                                2           33                3.3
                                                                3           50                  5
                           
         So,  k = 3,  x = 5,  and   A – xL1  =  A – 5L1,  and we have a
                            

                           Division Rule for 17:    If A – 5L1 is divisible by 17, then so is N.                   

 

     Example.    650,998  =  650,998    →    65099 – 5(8)  =  65059                             650998
                                                                                                                                              65099
                                                  65059   →       6505 – 5(9)  =    6460                                   -40
                                                                                                                                               65059
                                                    6460    →         646 – 5(0)  =      646                                 -45
                                                                                                                                               6460
                                                       646    →           64 – 5(6)  =        34                                 -0
                                                                                                                                               646
      Since 34 is divisible by 17, so is 650,998.    (650,998 ÷ 17 = 38,294                   -30
                                                                                                                                               34
 
                                                                Division Rule for 18
 
         Since 18 = 2(9), any natural number N divisible by 18 must be divisible by both 2 and 9. Thus, it must
    be even, and the sum of its digits must be divisible by 9.  Thus, we have the

                    

                          Division Rule for 18
 
                          If N is even, and the sum of its digits is divisible by 9, then N is divisible by 18.
                         ____________________________________________________________
    
     Example.    Since 13,652,874 is even and the sum of its digits, 36, is divisible by 9, then it is divisible by
    18.   In fact, 13,652,874 ÷ 18  =  758,493
 
 
                                                                       Division Rule for 19
 
          To find a Division Rule for 19, we use the General Division Rule above with D = 19, and we must find k
      so that x = (Dk – 1)/10 = (19k – 1)/10 is an integer. To find k, we construct the table below. 
 
                                                                  k       19k – 1     x = (19k – 1)/10
                                                                -----------------------------------------------
                                                                  1           18                 1.8
                                                                  2           37                 3.7
                                                                  3           56                 5.6
                                                                 -   -   -   -   -   -   -   -   -   -   - 
                                                                  9         170                 17
                                                    
              So, k = 9,  x = 17,  and   A – xL1  =  A – 17L1.  To get a simpler rule, let’s add  19L1  to
        A – 17L1,  getting A + 2L1  as a new rule for division by 19.
     
                      Division Rule for 19:    If A + 2L1 or A – 17L1 is divisible by 19, then so is N.        

 

     Example.   Is 1,353,617 divisible by 19?
 
                                                        A + 2L1                                       A – 17L1       
 
                                                      1353617                                     1353617
                                                            +14                                            -119
                                                     135375                                       135242
                                                          +10                                              -34
                                                      13547                                         13490
                                                        +14                                                -0
                                                      1368                                           1349
                                                      +16                                            -153
                                                      152                                              -19
                                                      +4
                                                      19
                                                                                                                          
        Since 19 (or -19) is divisible by 19, so is 1,353,617.    (1,353,617 ÷ 19  =  71,243) 
 
           
                                                                           Division Rule for 20
 
          Since 20 = 4(5) = 2(10), any natural number N divisible by 20 must be divisible by  2, 4, 5, and 10.
     Thus, N must be even,  the number formed by its last two digits must be divisible by 4,  its last digit must
      be 0 or 5 (but not 5, since N must be even),  and its last digit must be 0. Thus, we have the

 

                                       Division Rule for 20

 
                                        If the number formed by the last two digits of N is divisible by 4, 
                                        and its last digit is 0, then N is divisible by 20.
                                       _________________________________________________ 
      
     Example.  19,974,460 is divisible by 20, since the number formed by its last two digits, 60, is divisible
     by 4, and its last digit is 0.   In fact, 19,974,460 ÷ 20  =  998,723.
 
                                                                          Division Rule for 21
 

           Since 21 = 3(7), any natural number N divisible by 21 must be divisible by both 3 and 7. Thus, the

      sum of its digits must be divisible by 3, and A – 2L1 must be divisible by 7.

                                 Division Rule for 21

                                  If the sum of the digits of N is divisible by 3, and A – 2L1 is divisible
                                  by 7, then N is divisible by 21.
                                 _____________________________________________________
  
     Example.  1,233,456 is divisible by 21, since the sum of its digits, 24, is divisible by 3, and A – 2L1, as
     shown below, is divisible by 7.  In fact,  1,233,456  ÷  21  =  58,736.
 
                                                                               1,233,456
                                                                                        - 12
                                                                                123,333
                                                                                        - 6
                                                                                12,327
                                                                                   - 14   
                                                                                1,218
                                                                                 - 16 
                                                                                 105
                                                                               - 10
                                                                                    0,      which is divisible by 7.
 
                                                                        Division Rule for 22
 
         Since 22 = 2(11), any natural number N divisible by 22 must be divisible by both 2 and 11. Thus,
     it must be even, and A – L1 must be divisible by 11.
                 
           Division Rule for 22: If N is even, and A – L1 is divisible by 11, then N is divisible by 22.
                        
     Example.   1,925,814 is divisible by 22 since it is even, and A – L1, as shown below, is divisible by 11.
     In fact, 1,925,814 ÷ 22  =  87,537.
 
                                  1,925,814         192,577          19,250           1,925           187
                                             - 4                   - 7                  - 0                - 5               -7
                                   192,577           19,250             1,925              187              11,    divisible by 11.        
                                                        
                                                        
         Now, the General Division Rule for Any Prime Divisor D gives us rules of division involving just the last
     digit, L1, of a number N. Let’s expand this rule so that, in addition to this last digit, we may use the last two
     digits or more of N in finding new division rules.
                                                 
               
                                                  Expanded Division Rule for Any Prime Divisor D
 
     Last Two Digits of N
 
           Let’s write N as N = 100A + L2, where A is the number formed by ALL of the  digits of N except its
      last two, and L2 is the number formed by the last two digits of N.
 
           Then, N = 100A + L2.    Solving this for L2, we get L2 = N – 100A.                                                    (3)
 
      Now,      N  =  100A + L2  =  99A  +  L2  +  (A – xL2)  +  xL2,  where x is a positive integer.   Then,  
  
                     N  =    99A  +  (x + 1)L2  +  (A – xL2).
 
      Substituting for L2 from (3),   N     =    99A  +  (x + 1)(N – 100A)  +  (A – xL2)
 
                                           N  –  99A    =   (x + 1)N  –  100A(x + 1)  +  (A – xL2)
 
                           100A(x + 1)  –  99A   =   (x + 1)N   –   N   +   (A – xL2)
 
                                         100Ax  +  A   =    xN  +  N – N  +  (A – xL2)
 
                    (100x + 1)A  –  (A – xL2)   =   xN       
 
         Now, if we choose x so that  100x + 1  and  A – xL2  are both divisible by D,  then the left-hand side of
    the above equation will be divisible by D, and thus the right-hand side,  xN,  will be also. Now, if x were
    divisible by D, then 100x would also be divisible by D, and thus 100x+1 could not be divisible by D. But
     this contradicts the fact that we chose x so that 100x + 1 would be divisible by D.  So x cannot be divisible
     by D, and N must be divisible by D, since D is prime.
 
         Thus, if A – xL2 is divisible by D, where x is chosen so that 100x+ 1 is divisible by D, then N will be
      divisible by D.
 
         Note:  If 100x + 1 is divisible by D, then it is a multiple of D, so 100x + 1 = kD for some integer k. Solving
      for x, we get   x = (Dk – 1)/100.
 
          Furthermore, once we find A – xL2, we can find other division rules for D by adding multiples of DL2 to
       A – xL2, because for any integer k1, k1DL2 is divisible by D,  and if A – xL2 is divisible by D, then so is
       (A – xL2)  +  k1DL2.  All this gives us the
 

                  Last Two Digits Division Rule for Any Prime Divisor D  

   
                  Let N = 100A + L2, where A is the number formed by ALL of the digits of N  except its last two,
                  and L2 is the number formed by the last two digits of N.  Then, N is divisible by  D  if  A – xL­2­
                  is divisible by D, where D is prime and  x is an integer selected so that 100x + 1 is divisible
                  by D;  that is,  x  =  (Dk – 1)/100  for some integer k.
 
                  Furthermore, if A – xL2 is divisible by D, then so is   A – xL2 + k1DL2 for any integer k1.
                 _________________________________________________________________________

 

      If we generalize the Last Two Digits Division Rule to the last n digits of N, we get
 
 

                  Last n Digits Division Rule for Any Prime Divisor D

   
                  Let N = 10nA + Ln, where A is the number formed by ALL of the digits of N except its last n
                 digits, and Ln is the number formed by the last n digits of N. Then, N is divisible by D if A – xL­n
                  is divisible by D, where D is prime and  x is an integer selected so that 10nx + 1 is divisible by
                  D; that is,  x  =  (Dk – 1)/10n  for some integer k.
        
                  Furthermore, if A – xLn is divisible by D, then so is   A – xLn + k1DLn for any integer k1.
                 _________________________________________________________________________
 
                                                                               Division Rule for 23
 
          Let’s use the Last n Digits Division Rule above to find division rules for 23 which use a) the last digit
     of N,  b) the last 2 digits of N,  and c)  the last 3 digits of N.
  
        a) For the last digit of N, we have n = 1, D = 23, and we must find the value of k so that x = (Dk – 1)/10n
             =  (23k – 1)/101  =  (23k – 1)/10  is an integer.    Constructing the table below, we get
                                                                  k       23k – 1         x = (23k – 1)/10
                                                                -------------------------------------------------
                                                                  1            22                    2.2
                                                                  2            45                    4.5
                                                                  7          160                    16
                                                      
             So, k = 7, x = 9, and we ge t A – xLn  =   A – 16L1 for a division rule for 23. To get a simpler rule,
             we add  23L1  to  A – 16L1,  getting  A + 7L1  as a new rule for division by 23.
       
                                   Division Rule for 23  (a)  If A + 7L1 is divisible by 23, then so is N.        

 

        b)  For the last 2 digits of N, we have n = 2, D = 23, and we must find the value of k so that
              x  =  (Dk – 1)/10n  =  (23k – 1)/102  =  (23k – 1)/100  is an integer. Constructing the table
              below, we get
                                                                 k       23k – 1         x = (23k – 1)/100
                                                              ----------------------------------------------------
                                                                 1            22                     0.22
                                                                 7            90                     0.9
                                                               77      1771 – 1                1.77
                                                               87      2001 – 1                 20
                          
              So, k = 87, x = 20, and we get  A – xLn  =   A – 20L2  for a division rule for 23.  To get a simpler rule,
               we add  23L2  to  A – 20L2,  getting  A + 3L2  as a new rule for division by 23.
 
                                    Division Rule for 23  (b)  If A + 3L2 is divisible by 23, then so is N.        
 
        c)  For the last 3 digits of N, we have n = 3, D = 23, and we must find the value of k so that
              x  =  (Dk – 1)/10n  =  (23k – 1)/103  =  (23k – 1)/1000  is an integer. Constructing the table
              below, we get
 
                                                                  k       23k – 1        x = (23k – 1)/1000
                                                                ----------------------------------------------------
                                                                  1             22                    0.022
                                                                  7        161 – 1                0.16
                                                                77      1771 – 1               1.77
                                                                87      2001 – 1                   2
                        
               So, k = 87, x = 2, and we get  A – xLn  =   A – 2L3  for a division rule for 23. 
                  
                            Division Rule for 23 (c) If A – 2L3 is divisible by 23, then so is N. 
    
         Summarizing all this, we have
                           
                  Division Rule for 23:  If A + 7L1,  A + 3L2,  or  A – 2L3  is divisible by 23, then so is N.
 
     Example.    Is 780,252 divisible by 23? 
 
                          Using  A – 2L3,   we get        780252   →  780 – 2(252)  =  780 – 504  =  276
 
                          Using  A + 7L1,    we get              276   →     27 + 7(6)      =     27 + 42   =   69.
 
              Since 69 is divisible by 23, so is 780,252.
 
                                                                           Division Rule for 24
          Since 24 = 3(8),  any natural number N divisible by 24 must be divisible by both 3 and 8. Thus,
     the sum of the digits of N must be divisible by 3, and the number formed by its last 3 digits must be
     divisible by 8.  This gives us the   
                               Division Rule for 24 
                                      
                               If the sum of the digits of a natural number N is divisible by 3, and the number
                               formed by its last 3 digits is divisible by 8, then N is divisible by 24.
                              ____________________________________________________________  
       
     Example.  20,782,632 is divisible by 24 because the sum of its digits, 30, is divisible by 3, and the
      number formed by its last 3 digits, 632, is divisible by 8.  To show this, we divide 632 by 2 three times,
      getting   632 ÷ 2  =  316,  316 ÷ 2  =  158, and  158 ÷ 2  =  79. 
                        
                                                                            Division Rule for 25
        To find a division rule for 25, let’s write down the expression for any 6-digit number;
                                              N  =          a5105    +    a4104      +    a3103   + a2102    +  a1101  +  a0
                                                   =    100,000a5  + 10,000a4   + 1,000a3  + 100a2    +  10a1   +  a0
                                                   =   25(4,000)a5 + 25(400)a4 + 25(40)a3 + 25(4)a2  +  (10a1 + a0)   
      As we can see, every term except the last two (10a1 + a0) has a factor of 25, so 25 is a divisor of each
    term of N except the last two, 10a1 + a0. Now, if 25 is a divisor of 10a1 + a0, then every term of N will be
    divisible by 25, and N will be divisible by 25.  
               In general,               N   =    an10n   +   an-110n-1  +  .  .  .  +   a1101  +   a0 
                                =         2n5nan  +   2n-1 5n-1an-1   +   .  .  .   +   2353a3   +   2252a2  +  2(5)a1  +  a0                 
                                =        5n(2nan) +  5n-1(2n-1an-1)  +   .  .  .   +   53(23a3)  +  52(22a2)  +  (10a1 + a0),
    which shows that every term of N has a factor of 52 = 25 (for n > 1) except (10a1 + a0). Now, if 25 is a
    divisor of (10a1 + a0),  then every term of N will be divisible by 25, and thus N will be divisible by 25,
     giving us the

                             Division Rule for 25

                             If the number formed by the last two digits of a natural number N is divisible
                             by 25, then N is divisible by 25.   Note: The only two-digit numbers divisible
                             by 25 are  00, 25, 50,  and  75.
 
     Example.  37,475 is divisible by 25 because the number formed by its last two digits, 75, is divisible by
     25.   In fact  37,475 ÷ 25  =  1,499.
  
 
                                 Division Rules of the Form A – xLn for Selected Divisors
           
         If  N = 10A + L1,  N = 100A + L2,  or    N = 1000A + L3,   then N is divisible by D if any of the
         expressions opposite D in the table below are divisible by D.
  
                              D                                          Expressions Divisible by D
                     ------------------------------------------------------------------------------------------------------
               
                   3                                      A + L1                 A + L2                A + L3                          
                                                A – 2L1               A – 2L2              A – 2L3
 
                               7                                  A – 2L1               A – 3L2              A – L3                                  
                                                                 3A + L1                 2A + L2         
                     
                    9                                                                    A + L1                  A + L2               A + L3
                                           A – 8L1                A – 8L2             A – 8L3
 
                              11                                  A – L1                  A + L2               A – L3
                       
                              13                                  A – 9L1               A – 10L2            A – L3            
                                                                    A + 4L1                A + 3L2
     
                              17                                   A – 5L1               A – 9L2            A – 6L3    
 
                              19                                   A +  2L1               A + 4L2            A + 8L3
                                                                     A – 17L1              A – 15L2          A – 11L3
 
                              23                                    A + 7L1                A +  3L2           A – 2L3
 
                   Divisibility Rules for Natural Numbers for Divisors from 1 to 25
 
       For the rules below, N = 10A + L1, N = 100A + L2, or N = 1000A + L3, where
 
                    L1 is the last digit of N, and A is the number formed by the remaining digits of N,
 
                    L2 is the number formed by the last two digits of N, and A is the number formed
                            by the remaining digits of N, and
 
                   L3 is the number formed by the last three digits of N, and A is the number formed
                           by the remaining digits of N,
  
      Divisor                                                                    Rule
      
          1        Any number is divisible by 1.
 
    2        Any number whose last digit is even is divisible by 2.
 
    3        If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3.
 
    4        If the number formed by the last two digits of a number N is divisible by 4, then N is divisible by 4.
   
          5    Any number whose last digit is 0 or 5 is divisible by 5.
 
          6    Any number whose sum of its digits is divisible by 3 and whose last digit is even is divisible by 6.
 
          7     If N = 10A + L1, where A is the number formed by ALL of the digits of N except its last one,
                and L1 is the last digit of N, then N is divisible by 7 if A – 2L1 is divisible by 7.
                (See Division Rules for 7 above for more rules.)   
 
    8        If the number formed by the last three digits of a number N is divisible by 8, then N is divisible by 8.
 
    9    If the sum of the digits of a number is divisible by 9, then the number itself is divisible by 9.
 
         10   Any number whose last digit is 0 is divisible by 10.     
                      
         11   If  A – L1,  A + L2,  or  A – L3  is divisible by 11, then N is divisible by 11.
 
     12      Any number whose sum of its digits is divisible by 3 and whose last two digits are divisible
              by 4 is divisible by 12.
 
          13    If  A – 9L1,  A + 4L1,  A + 3L2,  A – 10L2,  or  A – L3  is divisible by 13, then so is N.   
              
      14    If N is even and A – 2L1 is divisible by 7, then N is divisible by 14.     
         
          15    If the last digit of N is 0 or 5, and the sum of its digits is divisible by 3, then N is divisible by 15.
    
          16    If the number formed by the last four digits of a natural number N is divisible by 16,  then N is
                  divisible by 16.
   
          17    If A – 5L1 is divisible by 17, then so is N.  
   
          18     If N is even, and the sum of its digits is divisible by 9, then N is divisible by 18.
    
          19     If  A + 2L1  or  A – 17L1  is divisible by 19, then so is N.   
       
          20     If the number formed by the last two digits of N is divisible by 4, and its last digit is 0, then N is
                   divisible by 20.
       
          21    If the sum of the digits of N is divisible by 3, and A – 2L1 is divisible by 7, then N is divisible by 21.
     
          22    If N is even, and A – L1 is divisible by 11, then N is divisible by 22.
       
          23    If A + 7L1,  A + 3L2,  or  A – 2L3  is divisible by 23, then so is N.
        
          24     If the sum of the digits of a natural number N is divisible by 3, and the number formed by its
                    last 3 digits is divisible by 8, then N is divisible by 24.  
     
          25     If the number formed by the last two digits of a natural number N is divisible by 25, then N
                   is divisible by 25.   Note: The only two-digit numbers divisible by 25 are  00, 25, 50,  and  75.
          
                                                                                                                                   January 11, 2008
 ___________________________________________________________________________________