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no, but you can get a fractional answer if you divide.

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Sum the digits in blocks of three from right to left. If the result is divisible by 27, then the number is divisible by 27

Well, because it has to be a multiple of 10, you know that it has to end with a 10. The 27 is the biggest number and will be what we focus on. To get the 27 into a multiple of 10, we have to multiply it by 10.10 x 27 = 270.270 is not divisible by 12 or 8, so we continue.Keeping in mind that the number has to be a multiple of 10 and 27, we can only add intervals of 270.270+270 = 540.540 is divisible by 27, 10 and 12, but not by 8. So lets add another 270.540+270 = 810810 is only divisible by 10 and 27, much like 270, so lets add another 270.810+270 = 10801080 is divisible by 27, 10, 12 and 8 and is the lowest common multiple.

9

Yes, it is 27

972 is one of the numbers that is divisible by both.

1,215

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.

How about 27

81 is divisible by 1, 3, 9, 27 and 81

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.

1080

27, 21

The easiest way to know if a number is divisible by 9 is to add the digits in the number. 4 + 7 + 8 + 8 = 27 Since 9 will go into 27 evenly, 4788 is divisible by 9.

you must add 10 368+10=378 378/27=14 If you divide 368 by 27, you get 13 and remainder 17. So you add 10.

77922 is divisible by 6. 77922 / 6 = 12987 To check if a number is divisible by 6, the number has to be even and divisible by three. An easy way to check if a number is divisible by three is to see if the sum of the digits is divisible by three. 77922 7+7+9+2+2 =27 27 2+7 =9 9 is divisible by three, so 77922 is divisible by three also.

Yes it is. In order for it to be divisible by 9, the sum of the digits in the number must also be divisible by 9. In this case - the sum of the digits is 27.

Well, it's not. Take 27 for example - divisible by 9 but not by 6.

yeah3 9 15 21 27 and so on(every other multiple of three)But if you have a large number, like 755253, and don't have a calculator handy, then use sum of digits to determine if divisible by 3.Then if the number is divisible by 3 and divisible by 2, then it is also divisible by 6.So in this case: 7 + 5 + 5 + 2 + 5 + 3 = 27, which is divisible by 3, but the original number is odd, so the number is not divisible by 6.

This is 1 over 9. You just have to find out what number both 3 and 27 are divisible by and then divide each number by that until you have no other number both the numerator and denominator are both divisible by (they both have to be divisible by the same number to make it equivalent).

27 is divisible by 1, 3, 9, 27.

Neither 494733 nor 27 is a prime number because they each are divisible by 3.

No. 135 is divisible by: 1, 3, 5, 9, 15, 27, 45, 135.

27 is wholly divisible by 1, 3, 9, and 27.

NO, it is not divisible by all numbers. This is easy to see if you remember that 27 is an odd number and odd numbers are NOT divisible by 2. Then there is the issue that is is NOT divisible by any number greater than itself. That is a lots of numbers that don't work!

If the number has a remainder of 2 when divided by 5, it must end in either 2 or 7. Because adding 10 to a number ending in 2 will not be a prime number, it must end in 7. Because you will have a prime number if you subtract 10, it must be greater than 10. The possible numbers are: 17 27 37 47 The only one of these that is divisible by 3 is 27. If you subtract 10 from 27, you have 17, which is a prime number. If you add 10 to 27, you have 37, which is a prime number. The factors of 27 are 1, 3, 9, and 27, so it has only two factors other than 1 and itself. When divided by 5, the result is 5 remainder 2. Thus, 27 fits all the conditions.