Any multiple of 81.
They're not. 81 is divisible by 1, 3, 9, 27, and 81. Those are called the "factors" of 81, because each of them divides evenly into 81 with no remainder.
There are only three proper factors of 81; they are 3, 9 and 27.81 is also divisible by 1 and by itself, but mathematicians regard these as improper factors because they are trivial. Every natural number is divisible by itself and by unity.
"What is divisible by 81?" Any multiple of 81 is divisible by 81. For example, all of the following are divisible by 81: 81 = 1 x 81 162 = 2 x 81 243 = 3 x 81 324 = 4 x 81 and so on. ========================================= Did you mean to ask "What is 81 divisible by?" 81 is divisible by 1, 3, 9, 27, 81.
81 is divisible by 1, 3, 9, 27 and 81
The factors of 81 are: 1, 3, 9, 27, 81 The factors of 150 are: 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150
81 has factors other than 1 and itself, so it is not a prime number. For example, it is divisible by 3. (You can check this by adding the digits of the number: 8 + 1 = 9. If the sum of the digits is divisible by 3, the number is also divisible by 3.) Therefore, 81 is a composite number.
No, 81 is not divisible by 4. In order for a number to be divisible by 4, the last two digits must be divisible by 4. In this case, the last two digits of 81 are 81 itself, which is not divisible by 4. Therefore, 81 is not divisible by 4.
It has more than two distinct factors; they are 1, 3, 9, 27, 81
Techcially all numbers are divisible by 81, however many would result in a decimal answer. If you mean how many numbers are divisible by 81 and have an integer answer, then it is "all multiples of 81" - e.g 81 x 2, 81 x 3 etc...81, 162, 243 and keep adding 81 forever.
No, it's not. 243 is divisible by 3 and 81. To be prime, a number's only factors must be 1 and itself.
no because 81 is an odd number. only even numbers are divisible by 2 evenly.
9 x 9 = 81