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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
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