yes the answer to 1 divided by 3 = 1/3 (this is 1 third)
Oh, dude, numbers divisible by both 3 and 9 are multiples of the least common multiple of 3 and 9, which is 9. So, any number that is a multiple of 9 is also divisible by 3 and 9. Like, it's just basic math, nothing to lose sleep over.
One-third of them
To determine if 4908 is divisible by 3, we need to sum its digits. 4 + 9 + 0 + 8 = 21. Since 21 is divisible by 3 (21 ÷ 3 = 7), 4908 is also divisible by 3. Therefore, 4908 is divisible by 3.
Well, darling, the positive integers divisible by 2 are all the even numbers, and the ones divisible by 3 are those multiples of 3. So, to find the numbers divisible by either 2 or 3, you just need to combine these two sets. In other words, it's the set of all even numbers and multiples of 3. Voilà!
yes.
No, just every third one.
At least one of the four will be divisible by 4. One other will be divisible by 2 (but not by 4) One other will be divisible by 3. Hence the product will be divisible by 4 x 2 x 3 = 24.
No, it is divisible by 3.No, it is divisible by 3.No, it is divisible by 3.No, it is divisible by 3.
3 is one factor.
34 is divisible by 2 seventeen times. 34 is divisible by 3 eleven times with one left over.
One way to work it out in your head is to consider than 30 is divisible by 3, so you can ask if 140 is divisible by 30. It's not, since 14 is not divisible by 3. If the number was 129 or 159, both of those are divisible by three since both 12 and 15 are divisible by 3.
No, this number is divisible by 3 and hence is not a prime number. You can tell that fast by adding the individual digits. If that sum is itself divisible by 3, as this one is, then the number itself is divisible by 3.
To be divisible by 15, it must also be divisible by 3 and by 5. To be divisible by 3, the sum of the digits must be divisible by 3; to be divisible by 5, the number must end with a zero or a five. Considering all these criteria, I guess that number would be 1110.
Yes, if x is an integer divisible by 3, then x^2 is also divisible by 3. This is because for any integer x, x^2 will also be divisible by 3 if x is divisible by 3. This can be proven using the property that the square of any integer divisible by 3 will also be divisible by 3.
A number is divisible by 3 if the sum of its digits is divisible by 3.
Let the three integers be, n, (n + 1), and (n + 2) Then at least one of these numbers is even and therefore has a factor of 2. And one of the numbers is divisible by 3 **. Therefore the product has factors of 2 and 3 and is thus divisible by 2 x 3 = 6. ** Either n is divisible by 3. Or, n leaves a remainder of 1 when divided by 3 in which case (n + 2) is divisible by 3. Or, n leaves a remainder of 2 when divided by 3 in which case (n + 1) is divisible by 3.
Check your divisibility rules: 2: One's digit is an even number YES 3: The sum of the digits is divisible by 3 NO: 7 + 1 + 0 = 8 which is not divisible by 3 5: One's digit is a 5 or a 0 YES 9: The sum of the digits is divisible by 9 NO: 7 + 1 + 0 = 8 which is not divisible by 9 10: One's digit is a 0 YES