It is always divisible by two.
True. Since 6 is divisible by 2, any number that is divisible by 6 will automatically be divisible by 2.
50% of numbers divisible by 5 are even, the other 50% are odd. If the single's digit of a number is 5, then the number is divisible by 5 and odd. For example, 25. However, if the single's digit of a number is 0, then the number is divisible by 5 and even. For example, 20.
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.
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.
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.
(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.
No. The reverse is true, but 12 is divisible by 4 and not by 8.
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
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
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.
Yes, it's even.
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.
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.
no 6 is divisible by 3 and so is 24 and so on ! lol
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.
Because it is divisible by a number other than 1 and itself. To be a prime number 58 would have to have no other divisors other than 1 and 58, which in this case is not true. 58 is divisible by 2 and 29 also, which makes it composite.
Here are some even numbers that are divisible by 5: 10 20 30 40 50 100 600 720 890 1110 2340 3360 108,070 224,560 Any number that is divisible by 10 (which means it ends in "0") is also divisible by 5.
False. An enormous number of them are divisible by three.
There does not exist a number that is divisible by all integers. The opposite is true. The number one can divided into all integers.