By using the divisibility rules, I can tell that 864 is divisible by 2, 3, 4, 6, 8 and 9. By dividing those numbers into 864 I can create factor pairs, any of which I can use to start the tree.
864
432,2
216,2,2
108,2,2,2
54,2,2,2,2
27,2,2,2,2,2
9,3,2,2,2,2,2
3,3,3,2,2,2,2,2
No you cannot. You might want to try the factor tree for 27.
No.
109 109 x 1 This is all there is to the factor tree, because 109 is a prime number. Normally, I wouldn't use 1 in a factor tree, but I do in this case to make it clear that I have done a factor tree, since 109 is only divisible by itself and 1.
IT IS PRIME there is no factor tree
The factor tree of 50:5025,25,5,25025,25,5,2
no
The factor tree is as follows: 114 57,2 19,3,2
No.
yes
991 is already prime. No tree is required.
13 is a prime number. It doesn't have a tree because its only prime factor is itself.
No you cannot. You might want to try the factor tree for 27.
15,3 3,5,3
prime factors
39 3,13
119 7,17
35 5,7