3*21 = 63 so 63 is divisible by 3
7*9 = 63 so 63 is divisible by 7.
Yes they are. They are both divisible by 7. 14/2= 7 and 63/9=7
9
Yes, 63 is divisible by 3. You can determine this by adding the digits of 63 (6 + 3 = 9), and since 9 is divisible by 3, so is 63. Additionally, dividing 63 by 3 gives a quotient of 21 with no remainder, confirming its divisibility.
Because 63 is evenly divisible by each one of the given numbers. Evenly divisible means the division operation yields zero remainder
126 is divisible by 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126.
1, 3, 7, 9, 21, 63
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.
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
1, 63, 3, 21, 7, 9
Yes they are. They are both divisible by 7. 14/2= 7 and 63/9=7
Yes beacause 9 times 7 is 63
63 ÷ 3 = 21
9
189 is divisible by: 1, 3, 7, 9, 21, 27, 63, 189.
Yes and the answer is 9 because 9*7 = 63
1, 3, 7, 9, 21, 49, 63, 147, 441