Prime divisors of 120 = 2, 4, 8, 3, 24, 5
The divisors of 63 are: 1, 3, 7, 9, 21, 63.
1, 3, 7, 9, 21, 63.
1, 3, 7, 9, 21, 63
1, 3, 7, 9, 21, 63.
293 only has two divisors because it is a prime number. The divisors are 1 and 293.
Since 7 is a prime, its only divisors are 1 and 7.
23 is a prime number. It has two divisors.
7 and 9 are divisors of 63.
3 is the prime number, because it has only itself and the number 1 as divisors. 4 and 10 have other divisors, such as 2, and are hence not prime.
yes. The only divisors of a prime number are itself and 1. Without counting the prime number, the sum of it's divisors is always 1.
29 is a prime number and so its only divisors are 1 and 29.
Because five is a prime number its only divisors are 1 and itself.
Since 11 is a prime number, its only divisors are 1 and itself.
13 is prime number therefore it only has two divisors 1 and 13
Any prime number to the 99th power has one hundred divisors.
Numbers called relatively prime or mutually prime have no common factors or divisors other than 1. Numbers having any common factors or divisors greater than 1 are not relatively or mutually prime.
Composite, for it has four positive integer divisors, 1, 3, 9, 27. To be a prime number, it needs to have only two integer divisors, 1 and itself.
Most certainly not. A prime is a number that has exactly two divisors: 1 and itself. Example: 2 is a prime. 4 is not; its divisors are: 1, 2 and 4.
No, 63 is not prime.
109 is a prime number. Therefore, the only divisors are 1and 109.
Co-prime or relatively prime.
47 does not have 4 prime numbers (factors or divisors).
Eight. The odd divisors of 756 are: 1, 3, 7, 9, 21, 27, 63, 189.
An integer (call it 'x') has exactly 3 divisors if and only if it is the square of a prime number. In other words, to generate a list of integers with exactly 3 divisors, just keep squaring prime numbers. A number with 3 divisors cannot be prime (a prime number has only 2 divisors, 1 and itself). So it must be a composite number, which is a number that can be factored as a product of prime numbers (Fundamental Theorem of Arithmetic) -- i.e. a composite number must have at least one prime divisor. In the case where the number has only 3 divisors, two of them are 1 and the number itself (neither of which are prime). Therefore the third divisor must be a prime number. So the three divisors of 'x' are: 1, p, x where p is prime. Now since p is a divisor (or factor) of x, and the only other divisor besides 1 and x itself, x must equal p*p -- or x=p^2 . Obvious x can't equal p*x and if x = p*1, x=p so x is prime, or has only 2 divisors... If x = p^(3) , then x = p*p* p , or p*(p^2) ... this means that p^2 would also have to be a divisor of x, and this would contradict with x having only 3 divisors. For the same reason, x = p^(greater than 3) is also not possible. So the only possibility is that an integer with exactly 3 divisors is the square of a prime number "p". The divisors are 1, p, and p^2. I'm sure there's a simpler, more elegant way of explaining this, but it should be clear enough.