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
119 7,17
39 3,13
No.
51 3,17