As 6 is divisible by 3, ANY dumber divisible by 6 is also therefore divisible by 3.
Any number divisible by 3 is ALSO a multiple of 3.
If a number is divisible by both 2 and 3, it is also divisible by 6. This is because 6 is the least common multiple of 2 and 3. Therefore, any number that meets the criteria of being divisible by both 2 and 3 will also be divisible by 6.
Yes, if a number is a multiple of 9, it is also divisible by 3. This is because 9 is itself a multiple of 3 (9 = 3 × 3), so any multiple of 9 can be expressed as 9n for some integer n, which can also be factored into 3(3n), demonstrating divisibility by 3. Therefore, all multiples of 9 are inherently divisible by 3.
No number that is a multiple of 3, can be a prime number. A prime number must only be divisible by itself and 1. It cannot be divisible by any other number. Therefore if it is a multiple of 3, then it must be divisible by 3 and hence, not a prime number.
It is a multiple of 6. There are an nfinite number of them and so cannot be listed. Each of these will be divisible by 6 (by definition). They will also be divisible by 2 and 3.
A number is divisible by 3 if the sum of its digits is a multiple of 3. A number is divisible by 6 if the sum of its digits is a multiple of 3 and it's even. A number is divisible by 9 if the sum of its digits is a multiple of 9.
A number that is divisible by 6 but not by 3 must be a multiple of 6 that is not a multiple of 3. Since 6 is a multiple of 3 (6 = 2 * 3), any multiple of 6 will also be a multiple of 3. Therefore, there is no number that is divisible by 6 but not by 3.
Because 9 is a multiple of 3.
If a number is divisible by both 2 and 3, it is also divisible by 6. This is because 6 is the least common multiple of 2 and 3. Therefore, any number that meets the criteria of being divisible by both 2 and 3 will also be divisible by 6.
Yes, if a number is a multiple of 9, it is also divisible by 3. This is because 9 is itself a multiple of 3 (9 = 3 × 3), so any multiple of 9 can be expressed as 9n for some integer n, which can also be factored into 3(3n), demonstrating divisibility by 3. Therefore, all multiples of 9 are inherently divisible by 3.
If a number is divisible by 4, it also means that the same number is divisible by 2. But if the number ends in a 3, it can't be divisible by 2 and, to a further extent, can't be divisible by 4.
No number that is a multiple of 3, can be a prime number. A prime number must only be divisible by itself and 1. It cannot be divisible by any other number. Therefore if it is a multiple of 3, then it must be divisible by 3 and hence, not a prime number.
It is a multiple of 6. There are an nfinite number of them and so cannot be listed. Each of these will be divisible by 6 (by definition). They will also be divisible by 2 and 3.
A number is divisible by 3 if the sum of its digits is a multiple of 3. A number is divisible by 6 if the sum of its digits is a multiple of 3 and it's even. A number is divisible by 9 if the sum of its digits is a multiple of 9.
If the number is even, it is a multiple of 2 If the sum of the digits make a number divisible by 3, the number is a multiple of 3 If the number ends in 5 or 0, the number is a multiple of 5 If the number is divisible by 2 and 3, the number is a multiple of 6 If the sum of the digits make a number divisible by 9, the number is a multiple of 9
If a number is even, it's divisible by two. If the digits of a number add up to a multiple of 3, it's a multiple of 3. For example, 342 is divisible by 3 because 3 + 4 + 2 = 9 and 9 is divisible by 3.
a 3 digit number that is divisible by on is a three digit number that is a multiple of one.
Any number that is divisible by 3 is a multiple of 3.