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3 and 9.

93 has a digit sum of 12, initially, which is divisible by 3, but not by 9. So 93 is divisible by 3, but not by 9.

99 has a digit sum of 18, initially, which is divisible by 3 and 9. So 99 is divisible by both 3 and 9.

Q: What is the number n where if the sum of the digits of the number is divisible by n the number itself is divisible by n?

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(The assumes that "the number" in the question is not n, although if they are they same number, this is still true.) "If the sum of the digits of the number is divisible by n, then the number itself is divisible by n" is true if n is 3 or if n is 9.

Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.

Yes.

The rule to decide if a number is divisible by 9 is to look at the sum of the digits. If the sum of the digits is divisible by 9, then so is the number.

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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That means the number itself is divisible by three. the sum of the digits in 144, for example, is 9. 144 divided by 3 is 48. The sum of the digits in 21 is 3. 21 divided by 3 is 7.If the number is an even number and the sum of the digits is divisible by 3, the number is also divisible by 6. 144 is even and the sum of the digits is divisible by 3, so 144 is also divisible by 6. 144/6 = 24. And finally, if the sum is divisible by 9, the number itself is also divisible by 9. 27 is an example of this.

(The assumes that "the number" in the question is not n, although if they are they same number, this is still true.) "If the sum of the digits of the number is divisible by n, then the number itself is divisible by n" is true if n is 3 or if n is 9.

This works with 3, and with 9.

That works with 3 and 9.

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

n = 1, 3 or 9.

If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.

Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.

No, this number is divisible by 3 and hence is not a prime number. You can tell that fast by adding the individual digits. If that sum is itself divisible by 3, as this one is, then the number itself is divisible by 3.

No. The sum of its digits, 5 + 3 + 1, is divisible by 3, which means the number itself is also divisible by 3.

The sum of the digits is 27, which is divisible by 9, so the number itself is divisible by 9. When divided by 9, it equals 20307.

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