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I'm not quite sure about that theory, but if all the numbers in the number add up to a number that's dividable by 3, then the number itself would then be divisible by 3. Ex. 141 1+4+1= 6 6 is divisible by 3, so 141 is too. 141/3= 47.

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Q: If every digit of a number is dividable by 3 then the number itself is divisible by 3?
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If every digit of a number is divisible by 3 then the number itself is divisible by 3?

Yes.


What largest 3 - digit number divisible by both 6 and 9?

999 is divisible by 9, but not by six; the next lower number divisible by 9 is 990, which is also divisible by 6, so that's the answer. Some shortcuts for divisibility: 0 is divisible by any number. If the last digit of a number is divisible by 2, the number itself is divisible by 2. If the sum of the digits of a number is divisible by 3, the number itself is divisible by 3. If the last TWO digits of a number are divisible by 4, the number itself is divisible by 4. If the last digit of a number is divisible by 5, the number itself is divisible by 5. If a number is divisible by both 2 and 3, it is divisible by 6. If the last THREE digits of a number are divisible by 8, the number itself is divisible by 8. If the sum of the digits of a number is divisible by 9, the number itself is divisible by 9. 990: 9+9+0=18, which is divisible by 9, so 990 is divisible by 9. 18 is also divisible by 3, so 990 is divisible by 3, and since 990 ends in 0 it's also divisible by 2, meaning that it's divisible by 6 as well.


What is the converse inverse and contrapositive If every digit of a number is divisible by 3 then the number is divisible by 3?

Converse:If a number is divisible by 3, then every number of a digit is divisible by three. Inverse: If every digit of a number is not divisible by 3 then the number is not divisible by 3? Contrapositive:If a number is not divisible by 3, then every number of a digit is not divisible by three.


What is a 4-digit number that is divisible by only 6 and 9?

There is no number that is divisible by only 6 and 9 because any number is divisible by 1 and itself. Furthermore, any number that is divisible by 6 MUST be divisible by 2 as well as 3 - so only 6 is impossible.


What is the rule for a number that is divisible by 4?

If the last two digits are divisible by 4 then the number is divisible by 4. Thus, if the tens digit is even and the units digit is 0 or 4 or 8 OR if the tens digit is odd and the units digit is 2 or 6 then the number is divisible by 4.