A number is divisible by 3 if the sum of its digits is a multiple of 3. For example, 27 = 2 + 7 = 9/3 = 3. Therefore, 27 is divisible by 3.If the sum of the digits of the number in question is divisible by three, the whole number is divisible by three.
No remainder. It has the same rule as 3 for divisibility. Add them up and if that is divisible by 27 then the number is divisible by 27.
Yes. Test a couple numbers divisible by 9--27, 81--and they are all divisible by 3.
54 is divisible by 2. Quotient is 27. 54 is divisible by 3. Quotient is 18. 54 is divisible by 9. Quotient is 6.
No - 3/81 = 0.037 recurring (that is, 0.037037037...)
27 is divisible by 1, 3, 9, 27.
27 is wholly divisible by 1, 3, 9, and 27.
No, the sentence structure of your sentence is wrong. It has to be "Is 27 divisible to 3?" Well, if you meant that, then yes, it is divisible. 27 divided by 3 is 9.
They are both divisible by 1 and 3
1 3 9 27.
A number is divisible by 3 if the sum of its digits is a multiple of 3. For example, 27 = 2 + 7 = 9/3 = 3. Therefore, 27 is divisible by 3.If the sum of the digits of the number in question is divisible by three, the whole number is divisible by three.
Yes, 27 is divisible by 1, 3, 9, 27.
Since the digits of 210 add up to 3, and that is divisible by 3 then 210 is divisible by 3. This works for 3, 9, and 27.
No remainder. It has the same rule as 3 for divisibility. Add them up and if that is divisible by 27 then the number is divisible by 27.
It's divisible by 3
Everything's divisible. 243 is divisible by 1, 3, 9, 27, 81 and 243.
Yes. Test a couple numbers divisible by 9--27, 81--and they are all divisible by 3.