18 is divisible by 1, 2, 3, 6, 9, and 18.
All multiples of 6 are evenly divisible by 6, such as: 6, -6, 12, -12, 18, -18, and so on. Numbers divisible by six must also be divisible by both 2 and 3.
Both. 18 / 6 = 3 and 18 / 9 = 2
Yes it is divisible by 2, 3, 6, and 9
Since the sum, 18, of the digits of 729 is divisible by 3, 18 = 3 x 6, then 729 is divisible by 3.
18 is divisible by 1, 2, 3, 6, 9, and 18.
All multiples of 6 are evenly divisible by 6, such as: 6, -6, 12, -12, 18, -18, and so on. Numbers divisible by six must also be divisible by both 2 and 3.
18
Not always because 18 is divisible by 2 and 6 but not by 12
Both. 18 / 6 = 3 and 18 / 9 = 2
No. Consider 6 itself or 18.
Yes it is divisible by 2, 3, 6, and 9
Numbers are divisible by 6 if they are even and also divisible by 3. Therefore, for example, since 18 is divisible by 3 and 2, it is also divisible by 6.
No; for example, 6 is not divisible by 12, nor is 18.
Since the sum, 18, of the digits of 729 is divisible by 3, 18 = 3 x 6, then 729 is divisible by 3.
First we have to find a prime number by which 18 is divisible. 18 is divisible by both 2 and 3. However, we can start with any number. 18/3 = 6, now we have to look for a prime number by which 6 is divisible. 6/3 = 2, 2 is divisible by itself. So, prime factorization of 18 = 2x3x3.
Yes, a number that is divisible by both 2 and 6 will always be divisible by 12. This is because 12 is the least common multiple of 2 and 6. Any number that is divisible by both 2 and 6 will also have all the factors of 12, making it divisible by 12 as well.