NO, lets take an example.. 24 is divided by 2 and 8 both.. but not by 16..
The number 32 is divisible by 1, 2, 4, 8, 16, and 32.
64 is divisible by 1, 2, 4, 8, 16, 32, 64.
16 is even because it is divisible by 2 and has a 6 in the ones place.
The number 2 is divisible by the multiples of 2 (which are infinite), including these: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.
Nope. 16 is divisible by 1, 2, 4, 8, 16.
yes it is because 16 is divisible by 2,4,and 8
NO, lets take an example.. 24 is divided by 2 and 8 both.. but not by 16..
6+8+2= 16, 16 isn't divisible by 3, so no, 682 is not
No. It doesn't end in a 5 or 0.
Not evenly because it will have a remainder of 4
Not evenly. 16 and 18 are.
8, 16, 24, 32... All multiples of 8 are also divisible by 2 and 4.
The digital root (sum of digit) must be divisible by 9, and the number formed by the last 4 digits must be divisible by 16. The second requirement ensures that the number is divisible by 16.
The number 32 is divisible by 1, 2, 4, 8, 16, and 32.
72 and 16 are divisible by 1, 2, 4, 8.
smallestnumber divisible by both 16 and 24 = 48Prime factorization of:16 = 2 * 2 * 2 * 224 = 2 * 2 * 2 .....* 3=============LCM=2* 2 * 2 * 2 * 3 = 48