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
2007-09-20
10:16:09
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2 answers
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asked by
MathTutor
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