Divisibility RulesWhy do these 'rules' work?  Dr. RobDivisibilidad por 13 y por números primos (13,17,19...)

Dividing by 3
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 3digit number with sequential digits, e.g., 123, 234, 456, is divisible by 3. Dividing by 4
Examples: Dividing by 5
Dividing by 6
Another easy way to tell if a [multidigit] 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
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.
If you do not know if a twodigit 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 threedigit 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 fourdigit number abcd, take the digit a and multiply by 6, then add 6a to bcd giving. This usually gives you a threedigit 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.
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.
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
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
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.
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. Twentyfour is divisible be eight. Therefore, our original number, one hundred and four, is also divisible by eight. Dividing by 9
Dividing by 10
Dividing by 11
Now look at 3531, which is also divisible by 11. It is not a coincidence that 3531 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 8Now write the numbers we saved in reverse order, and we have 8562415, which multiplied by 11 is 94186565.
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
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:

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