Both.
78 is evenly divisible by 1, 2, 3, and 13.
2, 3, 6, and 39
No,because 45 is not divisible by both 2 and 3. In order for a number to be divisible by 6 it has to be divisible by both 2 and 3.
If a number is divisible by both 2 and 3, it is also divisible by 6. This is because 6 is the least common multiple of 2 and 3. Therefore, any number that meets the criteria of being divisible by both 2 and 3 will also be divisible by 6.
The number 90 is divisible by both 2 and 3.
yes to both.
78 is evenly divisible by 1, 2, 3, and 13.
All multiples of 78 are divisible by it. For example ( 78 X 2, 156) or (78 x 3 , 234) etc.
2, 3, 6, and 39
No,because 45 is not divisible by both 2 and 3. In order for a number to be divisible by 6 it has to be divisible by both 2 and 3.
Any number that is divisible by both 2 and 3 is divisible by 6.
2 yes, 3 no.
If a number is divisible by both 2 and 3, it is also divisible by 6. This is because 6 is the least common multiple of 2 and 3. Therefore, any number that meets the criteria of being divisible by both 2 and 3 will also be divisible by 6.
The number 90 is divisible by both 2 and 3.
432 is divisible evenly by both 2 and 3
For a number to be divisible by 6, it has to be divisible by 2 and 3, because 2*3=6.In other words, we could divide the number we're factorising (78) by the prime factors of the number we're testing (6, which is 2*3). We start by dividing by 2, because we know 78 is even: 78/2=39. Then, we divide the new number by the other prime factor - 3. 39/3=13 13 is a whole number with no remainders, therefore we can say that 78 is divisible by 6, and therefore 6 is a factor of 78. (6*13=78)
78, 156, 234 and so on.