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For a number to be divisible by 105 it must be divisible by 3, by 5 and by 7. So, divisibility by 3 requires all three of the following to be satisfied:

  1. Sum the digits together. Repeat if necessary. If the answer is 0, 3, 6 or 9 the original number is divisible by 3.
  2. If the final digit of the number is 0 or 5, the original number is divisible by 5.
  3. Take the number formed by all but the last digit. From it subtract double the last digit. Keep going until there is only one digit left. If it is 0 or 7 then the original number is divisible by 7.

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More answers

a number can be divided by 2 evenly only if the number ends in the following numbers...0, 2, 4, 6, 8

a number can be divided by 3 evenly only if the sum of all the digits in the number is divisible by 3

a number can be divided by 4 if the numbers last two digits are divisible by 4

a number can be divided by 5 only if the number ends in the following... 0, 5

a number can be divided by 6 only if the number is divisible by two and three

a number can be divide by 7 only if the number can be divided by 7 (no way for 7)

a number can be divided by 8 only if the last two digits of the number is divisible by 8

a number can be divided by 9 only if the sum of all the digits in the number is divisible by 9

a number can be divided by ten only if it ends with the number zero

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Q: What is the Rules of divisibility?
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