If a number is even (divisible by 2) and divisible by 3, then it must also be divisible by 6.
If this is a T-F question, the answer is false. It is true that if a number is divisible by 6, it also divisible by 3. This is true because 6 is divisible by 3. However, the converse -- If a number is divisible by 3, it is divisible by 6, is false. A counterexample is 15. 15 is divisible by 3, but not by 6. It becomes clearer if you split the question into its two parts. A number is divisible by 6 if it is divisible by 3? False. It must also be divisible by 2. A number is divisible by 6 only if it is divisible by 3? True.
Yes, that is true.
Not always because 33 is divisible by 3 but not by 9
Three is a prime number and isn't divisible by any whole number * * * * * True, but irrelevant to the question. Any number that is divisible by 10 MUST be divisible by 5. Therefore there are no such numbers.
False. The question consists of two parts: - a number is divisible by 6 if it is divisible by 3? False. It must also be divisible by 2. - a number is divisible by 6 only if it is divisible by 3? This is true but the false part makes the whole statement false.
No, 9 is divisible by 3 and 9, but not six 3 x 9 = 27, also not divisible by 6
True. Since 6 is divisible by 2, any number that is divisible by 6 will automatically be divisible by 2.
Being divisible by 4, means that it will also be divisible by 2, so that doesn't tell you anything about divisibility by 8. But if you divide the number by 2, and this quotient is divisible by 4, then yes the original number is divisible by 8.
How can the following definition be written correctly as a biconditional statement? An odd integer is an integer that is not divisible by two. (A+ answer) An integer is odd if and only if it is not divisible by two
False. An enormous number of them are divisible by three.
The rule of thumb is that if the sum of the digits of the number equals a number that is divisible by 3, then the number is divisible by 3.For example,372: 3+7+2 = 12, which is a multiple of 3 (3x4=12)So, 3 is a factor of 372(which is true, as 3 x 124 = 372).124: 1+2+4 = 7, which is not a multiple of 3So, 3 is not a factor of 124(which is true, as 124 is not evenly divisible by 3, 124/3 = 41.33333)
A number that is divisible by 2 is not always evenly divisible by 6. It must also be divisible by 3, and must not be smaller than 6. The numbers 2, 4, 8, 10, 14, 16, 20, 22, 26, and 28 etc. are all divisible by 2 but not (evenly) by 6. The opposite is true, though. Any number divisible by 6 is also divisible by 2.
(The assumes that "the number" in the question is not n, although if they are they same number, this is still true.) "If the sum of the digits of the number is divisible by n, then the number itself is divisible by n" is true if n is 3 or if n is 9.
All multiples of the number 6 are also divisible by both 2 and 3.
Yes, it's true. 3 is one of the factors of 6, so any number that has six as a factor also has 3 as a factor.
Yes, it is true 558/3 = 186 To find out if a number is divisible by 3, add the digits; if the sum is divisible by 3, so is the number 5+5+8 = 18 which is divisible by 3
No, but the reverse is true.
yes always this is true' example 1,2,3 sum is 6 and is divisible by 3
Multiples of 9 and 6 are also divisible by three, the reverse is not true. 15 is divisible by 3, but not 6 or 9. 27 is divisible by 3 and 9, but not 6. 12 is divisible by 3 and 6, but not 9. 54 is divisible by 3, 6 and 9.
no 6 is divisible by 3 and so is 24 and so on ! lol
Numbers divisible by 6 will have at least one 2 and one 3 as prime factors because 2 x 3 = 6. The same is true for 2, 4 and 8. Numbers divisible by 4 are also divisible by 2. Numbers divisible by 8 are also divisible by 4 and 2.
There does not exist a number that is divisible by all integers. The opposite is true. The number one can divided into all integers.