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Divisibility Rules 
Dividing by 3
Why does the 'divisibility by 3' rule work? 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 if a+b+c is. This explanation also works to prove the divisibility by 9 test. It is originates from modular arithmetic ideas. Dividing by 4
Examples: Dividing by 5
Dividing 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.
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 6 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) + (6×1) = 16  not a multiple of seven. 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? 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. Dividing by 9
Dividing by 10
Dividing by 11
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.
Dividing by 12
