Yes because if you multiply 63 by 9 it will give you 567.
is 567 divisible by both 2 and 9
Only 234 is divisible by both 2 and 9.
443 + 567 = 1,010
1134 ÷ 2 = 567 Therefore, 567 + 567 = 1134.
Yes, 567 is divisible by 3. When you divide 567 by 3, you get an even quotient of 189.
Yes because if you multiply 63 by 9 it will give you 567.
is 567 divisible by both 2 and 9
567 is divisible by 9. 32 and 133 are not.
No, actually it isn't. 567 is divisible by 3.
1, 3, 9, 189, 567, 1701
Oh, dude, let me just crunch some numbers real quick for you. So, 567 divided by 4 equals 141.75. And since 141.75 isn't a whole number, you can't evenly divide 567 by 4. So, nope, 567 is not divisible by 4. Math can be a real party pooper sometimes, right?
For each given number, add the digits. If the sum comes to nine, then it is divisible by '9'. 234 ; 2 + 3 + 4 = 9 So is dividble by '9' 345 ; 3 + 4 + 5 = 12 = 1 + 2 = 3 Not divisible by '9' 567 ; 5 + 6 + 7 = 18 = 1 + 8 = 9 so is divisible by '9'.
composite number as it is divisible by 1, 3, 7, 9, etc
A prime number is a number with two and only two factors: 1 (also known as unity) and itself. A composite number is a number with more than two factors; in other words, it has at least one factor besides 1 and itself. 567 has factors other than 1 and itself, so it is not a prime number. For example, it is divisible by 3. (You can check this by adding the digits of the number: 5 + 6 + 7 = 18. If the sum of the digits is divisible by 3, the number is also divisible by 3.) Therefore, 567 is a composite number.
Only 234 is divisible by both 2 and 9.
1 and 567 are a factor pair of 567 since 1 x 567= 567 3 and 189 are a factor pair of 567 since 3 x 189= 567 7 and 81 are a factor pair of 567 since 7 x 81= 567 9 and 63 are a factor pair of 567 since 9 x 63= 567 21 and 27 are a factor pair of 567 since 21 x 27= 567