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.
The number 90 is divisible by both 2 and 3.
78, 156, 234 and so on.
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.
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)
432 is divisible evenly by both 2 and 3
The number 90 is divisible by both 2 and 3.
78, 156, 234 and so on.
1, 2, 3, 6, 13, 26, 39, 78