3*21 = 63 so 63 is divisible by 3 7*9 = 63 so 63 is divisible by 7.
63 ÷ 3 = 21
1, 3, 7, 9, 21, 63
Yes. But it is not completely divisible by 4 since you will have a remainder. 63 divided by 4 = 15 and 3/4th.
9, 3, 63, 189
Yes but it will have remainder of 3
Any multiple of 63 is divisible by it.
the factors of 63 are 3, 3 and 7. The natural numbers that can be multiplied to make 63 are : 1 * 63 3 * 21 9 * 7 7 * 9 21 * 3 63 * 1
Yes. The number 63 is divisible by 3 and by 7 and by 9. Since 3 x 3 x 7 = 63 as you can see, the number 63 is composite. A composite number is divisible by an integer (counting number) other than 1 and itself.
1008 is divisible by 63 and 48.
9/63 = (9÷3)/(63÷3) = 3/21 = (3÷3)/(21÷3) = 1/7 but as 6+3 = 9, 63 is divisible by 9, so: 9/63 = (9÷9)/(63÷9) = 1/7
not a prime number as itβs divisible by 1, 63, 3, and 21