So, this question comes in two parts:

Find the number of divisors of 189,720 that are (a) composite numbers, and (b) divisible by five.

I tried to solve the question by using a tree diagram, as I was shown in class.

(a)

189,720
4 x 47,430
6 x 7,605
15 x 37, 944

(Sorry if the diagram isn't as clear online, but it's shaped like a tree!)

Then, I did this:

(p+1)(q+1)(r+1)
= (4+1)(6+1)(15+1)
= 560 divisors

(b)

189,720
= 5 x 37,944

(p+1)(q+1)(r+1)
= (5+1)
= 6

I tried to follow the method we used in class, but I'd appreciate if someone could look over this and tell if it was right! Thank you!

How did yui decide to use 4, 6, and 15 as divisors? I don't know the "tree" method, but I would think that you need to determine the prime factors of 189720, and then use (p+1)(q+1)r+1)(s+1)... where p. q. r. s. .. are the quantity of each prime factor. Let me give an example: consider the number 72. Its prime factors are 2,2,2,3,3; so we use (p+1)(q+1) = 4*3 = 12. Thus there are twelve divisors of 72. Two of those divisors are prime (2 and 3), and one of those is the number 1 and one of those is 72 itself, so that means 12-4 = 8 divisors of 72 are composite numbers. They are: 4, 6, 8, 9, 12, 18, 24, and 36. I suggest you follow this same procedure for 189720.

**EDIT** OK, I think I know what the tree method is, and you used it incorrectly. Your first line is OK: 4 x 47430 = 189720. But the second line is incorrect, it should include both 2 x 2 = 4 and 6 x 7905 = 47430. Then the third line would be 2x3 = 6 and 15 x 527 = 7905. Then keep going: 527 is divisible by 17. If you complete the tree it would look like the attached. Note that there are 3 instances of prime number 2, two instances of prime number 3, and one each of prime numbers 5, 17 and 31.

Attachment 48617