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A number is divisible by 3 if the sum of its digits is a multiple of 3.

A number is divisible by 6 if the sum of its digits is a multiple of 3 and it's even.

A number is divisible by 9 if the sum of its digits is a multiple of 9.

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Q: How are the divisibility rules for 3 6 and 9 alike?
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Related questions

What is the rule for divisibility of 3?

It is 3 6 9


What is the divisibility rules of 4 and 6?

4- If the last two digits are divisible by 4, the whole number is divisible by 4. 6- If the number is even and also divisible by 3, it is divisible by 6.


Is 46 divisible by 6?

No Because, You Add The Digits = 4+6=10 So Its Not Check it In divisibility rules :)


Is 312 divisible by 3?

Yes. 312 / 3 = 104Because, using your divisibility rules you know that the sum of the numbers (3+1+2) is 6 which is a multiple of 3. threfor you can say, that is does devide by three.


What is the divisibility test for 15?

It is divisibility by 3 and divisibility by 5.Divisibility by 3: the digital root of an integer is obtained by adding together all the digits in the integer, with the process repeated if required. If the final result is 3, 6 or 9, then the integer is divisible by 3.Divisibility by 5: the integer ends in 0 or 5.


What is the test of divisibility for 6?

A number is divisible by 6 if the number is divisible by 2 AND 3.


What is a divisibility rule for 6?

A number has to be even and also divisible by 3.


How can you test for the divisibility by 6?

If the number is also divisible by 2 and 3


How can you use divisibility rules to make a factor tree for 864?

By using the divisibility rules, I can tell that 864 is divisible by 2, 3, 4, 6, 8 and 9. By dividing those numbers into 864 I can create factor pairs, any of which I can use to start the tree. 864 432,2 216,2,2 108,2,2,2 54,2,2,2,2 27,2,2,2,2,2 9,3,2,2,2,2,2 3,3,3,2,2,2,2,2


Can you tell if whether 432 is divisible by 2 3 4 5 6 9 10?

Yes, you can tell using the divisibility rules. The answers are yes for all but 5 and 10.


How divisibility rules can help you find the prime factorization?

Suppose you were trying to find the prime factorization of 123. You know that half of the divisors will be less than the square root. Since the square root is between 11 and 12, you only need to test 2, 3, 5, 7 and 11 as prime factors. If you know the rules of divisibility, you already know that 2 and 5 aren't factors and 3 is. It saves time.


What are the divisibility rules for the number two?

Any multiple of two must end in 0, 2, 4, 6 or 8.