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Q: Is every number divisible by 3 and 9 is also divisible by 6 is this true?

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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.

If a number is even (divisible by 2) and divisible by 3, then it must also be divisible by 6.

Yes

Yes. Every number is divisible by itself or 1.

Not always because 33 is divisible by 3 but not by 9

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.

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.

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.

True. Since 6 is divisible by 2, any number that is divisible by 6 will automatically be divisible by 2.

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.

All multiples of the number 6 are also divisible by both 2 and 3.

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

(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.

True fact.

true

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.

No, but the reverse is true.

no 6 is divisible by 3 and so is 24 and so on ! lol

There does not exist a number that is divisible by all integers. The opposite is true. The number one can divided into all integers.

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

The Rule for 4 is if the last 2 numbers are divisible by 4 (The tens and ones place) For example. 10759283740 is divisible, while 10759283738 is not, because 4 does not go into 38. The rule for 6 is if the number is even and also divisible by 3. If this is true, then the number is divisible by 6.

Yes, it is true that 2 is the only even prime number. All even numbers are evenly divisible by 2 (that is the definition of an even number). The number 2 is also divisible by 2, however, prime numbers, like all numbers, are evenly divisible by themselves, so that does not disqualify 2 from being a prime number.

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.