Divisibility. Question ment to Mathematicias?

Question

I was given this exercise but I think that something is not right.

I) Let a , b y c three natural numbers so that
3D( a , c ) = 5D( b , c ) (D = Greatest common divisor)

i ) Prove that a is a multiple of 5 and that b is a multiple of 3

This is obvious. I wrote this part because it's necessary to solve the rest of the exercise.

ii) Find N = "xyzy" (x, y, z and y are the digits) knowing that N is a multiple of 5 and that N is a multiple of 9 and that the number "xy" is equals to the number "zy" +10

iii) If 100 ( a + b) + (c / 3) where N is the number found in ii), a is a 2 digits number and a is a multiple of 3 , number of divisors of a = 8 and m( b , c ) = 5985 (m= least common multiple), then find a , b y c

I get two numbers N, 5040 and 9585 but I can't find any solution for the part iii. Perhaps I am missing something. I'd like some of you to tell me what the problem is

a is a multiple of 15 (a= 30 to me), so is c, and b is a multiple of 3

Answer 1

a=30, b=63, c=855, N=9585
I solved it in your previous question

Answer 2

I don't really understand iii. Are you saying 100(a+b) + (c/3) = N? And are you saying that "a" is the 2-digit number that's a multiple of 3, or are you saying that about the expression?

If it helps, notice that the prime factorization of 5985 = 3 * 3 * 5 * 7 * 19. This means these can be the only prime factors of b or c, and they all have to appear at least once.