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  Divisibility Rules

Why do these 'rules' work? - Dr. Rob

Divisibilidad por 13 y por números primos (13,17,19...)
- en español, de
la lista SNARK

    From the Archives of the Math Forum's Internet project Ask Dr. Math - our thanks to Ethan 'Dr. Math' Magness, Steven 'Dr. Math' Sinnott, and, for the explanation of why these rules work, Robert L. Ward (Dr. Rob).

    Dividing by 3

      Add up the digits: if the sum is divisible by three, then the number is as well. Examples:
      1. 111111: the digits add to 6 so the whole number is divisible by three.
      2. 87687687. The digits add up to 57, and 5 plus seven is 12, so the original number is divisible by three.

      Why does the 'divisibility by 3' rule work?

      From: Dr. Math
      To: Kevin Gallagher
      Subject: Re: Divisibility of a number by 3
      
      As Kevin Gallagher wrote to Dr. Math
      On 5/11/96 at 21:35:40 (Eastern Time),
      
      >I'm looking for a SIMPLE way to explain to several very bright 2nd 
      >graders why the divisibility by 3 rule works, i.e. add up all the 
      >digits; if the sum is evenly divisible by 3, then the number is as well.
      >Thanks!
      >Kevin Gallagher
      
      The only way that I can think of to explain this would be as follows:
      Look at a 2 digit number:
      
         10a + b = 9a + (a + b).
      
      We know that 9a is divisible by 3, so 10a + b will be divisible by 3
      if and only if a + b is.  Similarly, 
      
         100a + 10b + c = 99a + 9b + (a + b + c),
      
      and 99a + 9b is divisible by 3, so the total will be iff a + b + c is.
      
      This explanation also works to prove the divisibility by 9 test.  
      It clearly originates from modular arithmetic ideas, and I'm not sure if
      it's simple enough, but it's the only explanation I can think of.
      
      
      Doctor Darren,  The Math Forum
       Check out our web site -  http://mathforum.org/dr.math/
      

      Another visitor suggests this as an easier explanation, relying on decomposition and place value:

      We know that 9 is divisible by 3
      
        10 = 9 + 1
      
      and
      
        30 = 9*3 + 3
      
      Similarly,
      
      3000 = 999*3 + 3
      
      With this kind of decomposition in mind, examine any number;
      for example, 1235:
      
      1235 = 1000
              200
               30
             +  5
      
      Now,
      
      1000 = 999*1 + 1
       200 = 99*2  + 2
        30 = 9*3   + 3
         5 =         5
      
      Add these remainders:
      
      1 + 2 + 3 + 5 = 11
      
      Eleven is not divisible by three, which tells us that 
      our initial number, 1235, is not divisible by 3.
      
      Another interesting fact about 3 is that any 3-digit
      number with sequential digits, e.g., 123, 234, 456,
      is divisible by 3.
      

    Dividing by 4

      Look at the last two digits. If the number formed by its last two digits is divisible by 4, the original number is as well.
      Examples:
      1. 100 is divisible by 4.
      2. 1732782989264864826421834612 is divisible by four also, because 12 is divisible by four.

    Dividing by 5

      If the last digit is a five or a zero, then the number is divisible by 5.

    Dividing by 6

      Check 3 and 2. If the number is divisible by both 3 and 2, it is divisible by 6 as well.

      Robert Rusher writes in:
      Another easy way to tell if a [multi-digit] number is divisible by six . . .
      is to look at its [ones digit]: if it is even, and the sum of the [digits] is
      a multiple of 3, then the number is divisible by 6.
      

    Dividing by 7

      To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number.
      Example: If you had 203, you would double the last digit to get six, and subtract that from 20 to get 14. If you get an answer divisible by 7 (including zero), then the original number is divisible by seven. If you don't know the new number's divisibility, you can apply the rule again.

      Matthew Correnti describes this method:
      If you do not know if a two-digit number, call it ab, is divisible 
      by 7, calculate 2a + 3b. This will yield a smaller number, and if 
      you do the process enough times you will eventually -- if the 
      number ab is divisible by 7 -- end up with 7.
      
      You can use a similar method if you have a three-digit number abc: 
      take the digit a and multiply it by 2, then add it to the number bc, 
      giving you 2a + bc; repeat and reduce until you recognize the 
      result's divisibility by seven. With a four-digit number abcd, take 
      the digit a and multiply by 6, then add 6a to bcd giving. This 
      usually gives you a three-digit number; call it xyz. Take that x 
      and, as described previously, multiply x by two and add to yz 
      (i.e., 2x + yz). Again, repeat and reduce until you recognize the 
      result's divisibility by seven.
      

      Ahmed Al Harthy writes in:
      To know if a number is a multiple of seven or not, we can use also 
      3 coefficients (1 , 2 , 3). We multiply the first number starting 
      from the ones place by 1, then the second from the right by 3, 
      the third by 2, the fourth by -1, the fifth by -3, the sixth by -2, 
      and the seventh by 1, and so forth.
      
      Example: 348967129356876. 
      
      6 + 21 + 16 - 6 - 15 - 6 + 9 + 6 + 2 - 7 - 18 - 18 + 8 + 12 + 6 = 16 
      means the number is not multiple of seven.
      
      If the number was 348967129356874, then the number is a multiple of seven 
      because instead of 16, we would find 14 as a result, which is a multiple of 7. 
      
      So the pattern is as follows: for a number onmlkjihgfedcba, calculate
      
      a + 3b + 2c - d - 3e - 2f + g + 3h + 2i - j - 3k - 2l + m + 3n + 2o.
      
      Example:  348967129356874.
      
      Below each digit let me write its respective figure.
      
      3  4  8  9  6  7  1  2  9  3  5  6  8  7  4
      2  3  1 -2 -3 -1  2  3  1 -2 -3 -1  2  3  1
      
      (3×2) + (4×3) + (8×1) + (9×-2) + (6×-3) + (7×-1) + 
      (1×2) + (2×3) + (9×1) + (3×-2) + (5×-3) + (6×-1) + 
      (8×2) + (7×3) + (4×1) =  14 -- a multiple of seven.
      

      Another visitor observes:
      Here is one formula for seven...
      
      3X + L
      
      L = last digit
      X = everything in front of last digit.
      
      All numbers that are divisible by seven have this in common. 
      There are no exceptions.
      
      For example, 42: 3(4) + 2 = 14.
      Seven divides into 14, so it divides into 42.
      
      Next example, 105: 3(10) + 5 = 35.
      Seven divides into 35, so it divides into 105.
      
      Here is another formula for seven:
      
      4X - L
      
      When using this formula, if you get zero, seven or a multiple of seven, 
      the number will be divisible by seven.
      
      For example, 56: 4(5) - 6 = 14.
      Seven divides into 14, so it divides into 56.
      
      Next example, 168: 16(4) - 8 = 56.
      Seven divides into 56, so it divides into 168.
      
      Similarly:
      
      The formula for 2 is 2X + L
      The formula for 3 is 4X + L
      The formula for 4 is 6X + L
      The formula for 5 is 5X + L
      The formula for 6 is 2X + L and 4X + L -- in other words, the formulas for 2 and 3
                                                must work before the number is divisible by 6.
      The formula for 9 is X + L
      The formula for 11 is X - L
      The formula for 12 is 2X - L
      The formula for 13 is 3X - L
      The formula for 14 is 4X - L and 2X + L -- in other words, the formulas for 7 and 2 
                                                 must work before the number is divisible by 14.
      The formula for 17 is 7X - L
      The formula for 21 is X - 2L
      The formula for 23 is 3X - 2L
      The formula for 31 is X - 3L
      

      Sara Heikali explains this way to test a number with three or more digits for divisibility by seven:
      1. Write down just the digits in the tens and ones places.
      2. Take the other numbers to the left of those last two digits, 
      and multiply them by two.
      3. Add the answer from step two to the number from step one.
      4. If the sum from step three is divisible by seven, then the 
      original number is divisible by seven, as well. If the sum is 
      not divisible by seven, then the original number is not 
      divisible by seven.
      
      For example, if the number we are testing is 112, then
      1. Write down just the digits in the tens and ones places: 12.
      2. Take the other numbers to the left of those last two digits, 
      and multiply them by two: 1 × 2 = 2.
      3. Add the answer from step two to the number from step one: 
      12 + 2 = 14.
      4. Fourteen is divisible be seven. Therefore, our original 
      number, one hundred twelve, is also divisible by seven.
      

    See also Explaining the Divisibility Rule for 7 in the Dr. Math archives.

    Dividing by 8
      Check the last three digits. Since 1000 is divisible by 8, if the last three digits of a number are divisible by 8, then so is the whole number.
      Example: 33333888 is divisible by 8; 33333886 isn't.

      How can you tell whether the last three digits are divisible by 8? Phillip McReynolds answers:

      If the first digit is even, the number is divisible by 8 if the last two digits are. If the first digit is odd, subtract 4 from the last two digits; the number will be divisible by 8 if the resulting last two digits are. So, to continue the last example, 33333888 is divisible by 8 because the digit in the hundreds place is an even number, and the last two digits are 88, which is divisible by 8. 33333886 is not divisible by 8 because the digit in the hundreds place is an even number, but the last two digits are 86, which is not divisible by 8.

      Sara Heikali explains this test of divisibility by eight for numbers with three or more digits:
      1. Write down the units digit of the original number.
      2. Take the other numbers to the left of the last digit,
      and multiply them by two.
      3. Add the answer from step two to the number from step one.
      4. If the sum from step three is divisible by eight, then the 
      original number is divisible by eight, as well. If the sum is 
      not divisible by eight, then the original number is not 
      divisible by eight.
      
      For example, if the number we are testing is 104, then
      1. Write down just the digits in ones place: 4.
      2. Take the other numbers to the left of that last digit,
      and multiply them by two: 10 × 2 = 20.
      3. Add the answer from step two to the number from step one:
      4 + 20 = 24.
      4. Twenty-four is divisible be eight. Therefore, our original
      number, one hundred and four, is also divisible by eight.
      

    Dividing by 9

      Add the digits. If that sum is divisible by nine, then the original number is as well.

    Dividing by 10

      If the number ends in 0, it is divisible by 10.

    Dividing by 11

      Let's look at 352, which is divisible by 11; the answer is 32. 3+2 is 5; another way to say this is that 35 -2 is 33.

      Now look at 3531, which is also divisible by 11. It is not a coincidence that 353-1 is 352 and 11 × 321 is 3531.

      Here is a generalization of this system. Let's look at the number 94186565.

      First we want to find whether it is divisible by 11, but on the way we are going to save the numbers that we use: in every step we will subtract the last digit from the other digits, then save the subtracted amount in order. Start with

                9418656 - 5 = 9418651     SAVE 5
           Then 941865  - 1 = 941864      SAVE 1
           Then 94186   - 4 = 94182       SAVE 4
           Then 9418    - 2 = 9416        SAVE 2
           Then 941     - 6 = 935         SAVE 6
           Then 93      - 5 = 88          SAVE 5
           Then 8       - 8 = 0           SAVE 8
      Now write the numbers we saved in reverse order, and we have 8562415, which multiplied by 11 is 94186565.


      Here's an even easier method, contributed by Chis Foren:

      Take any number, such as 365167484.

      Add the first, third, fifth, seventh,.., digits.....3 + 5 + 6 + 4 + 4 = 22

      Add the second, fourth, sixth, eighth,.., digits.....6 + 1 + 7 + 8 = 22

      If the difference, including 0, is divisible by 11, then so is the number.

      22 - 22 = 0 so 365167484 is evenly divisible by 11.

      See also Divisibility by 11 in the Dr. Math archives.

    Dividing by 12

      Check for divisibility by 3 and 4.

    Dividing by 13


      Rafael Ando contributes:

      Instead of deleting the last digit and subtracting it ninefold from the remaining number (which works), you could also add the deleted digit fourfold. Both methods work because 91 and 39 are each multiples of 13.

      For any prime p (except 2 and 5), a rule of divisibility could be "created" using this method:

    1. Find m, such that m is the (preferably) smallest multiple of p that ends in either 1 or 9.
    2. Delete the last digit and add (if multiple ends in 9) or subtract (if it ends in 1) the deleted digit times the integer nearest to m/10. For example, if m = 91, the integer closest to 91/10 = 9.1 is 9; and for 3.9, it's 4.
    3. Verify if the result is a multiple of p. Use this process until it's obvious.

    Example 1: Let's see if 14281581 is a multiple of 17.
    In this case, m = 51 (which is 17×3), so we'll be deleting the last number and subtracting it fivefold.

    1428158 - 5×1 = 1428153
    142815 - 5×3 = 142800
    14280 - 5×0 = 14280
    1428 - 5×0 = 1428
    142 - 5×8 = 102
    10 - 5×2 = 0, which is a multiple of 17, so 14281581 is multiple of 17.

    Example 2: Let's see if 7183186 is a multiple of 46.
    First, note that 46 is not a prime number, and its factorization is 2×23. So, 7183186 needs to be divisible by both 2 and 23. Since it's an even number, it's obviously divisible by 2.
    So let's verify that it is a multiple of 23:

    m = 3×23 = 69, which means we'll be adding the deleted digit sevenfold.
    718318 + 7×6 = 718360
    71836 + 7×0 = 71836
    7183 + 7×6 = 7225
    722 + 7×5 = 757
    75 + 7×7 = 124
    12 + 7×4 = 40
    4 + 7×0 = 4 (not divisible by 23), so 7183186 is not divisible by 46.

    Note that you could've stopped calculating whenever you find the result to be obvious (i.e., you don't need to do it until the end). For example, in example 1 if you recognize 102 as divisible by 17, you don't need to continue (likewise, if you recognized 40 as not divisible by 23).
    The idea behind this method it that you're either subtracting m×(last digit) and then dividing by 10, or adding m×(last digit) and then dividing by 10.


    Jeremy Lane adds:

    It may be noted that while applying these rules, it is possible to loop among numbers as results.

    Example: Is 1313 divisible by 13?
    Using the procedure given we take 13×3 and obtain 39. This multiple ends in 9 so we add four-fold the last digit.

    131 + 4×3 = 143
    14 + 4×3 = 26
    2 + 4×6 = 26
    ...

    Example: Is 1326 divisible by 13?
    Using the procedure given we take 13×7 = 91. This is not the smallest multiple, but it does show looping. The smaller multiple does loop at 39 as well. There are some examples where we would still need to recognize certain multiples. So we subtract nine-fold the last digit.

    132 - 9×6 = 78
    7 - 9×8 = -65 (factor out -1)
    6 - 9×5 = -39 (again factor out -1)
    3 - 9×9 = -78 (factor out -1)

    This only occurs though if the number does happen to be divisible by the prime divisor. Otherwise, eventually you will have a number that is less than the prime divisor.


    And here's a more complex method that can be extended to other formulas:
    1 = 1 (mod 13)
    10 = -3 (mod 13)    (i.e., 10 - -3 is divisible by 13)
    100 = -4 (mod 13)   (i.e., 100 - -4 is divisible by 13)
    1000 = -1 (mod 13)  (i.e., 1000 - -1 is divisible by 13)
    10000 = 3 (mod 13)
    100000 = 4 (mod 13)
    1000000 = 1 (mod 13)
    
    Call the ones digit a, the tens digit b, the hundreds digit c, ..... and you get:

      a - 3×b - 4×c - d + 3×e + 4×f + g - .....

    If this number is divisible by 13, then so is the original number.

    You can keep using this technique to get other formulas for divisibility for prime numbers. For composite numbers just check for divisibility by divisors.

Dividing by 14

 

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8 November 2004