Thread: divisors

Could you please help me with this question:

Find all positive integers N such that the product 2005 *N has exactly six divisors.

5 * 401 = 2005

5*401*2*2*2*2 = 2005*16
5*401*2*2*2*3 = 2005*24
5*401*2*2*2*5 = 2005*60

this could go on forever. Ther are infinites N's where this would work.

Originally Posted by Math Help

5 * 401 = 2005

5*401*2*2*2*2 = 2005*16
5*401*2*2*2*3 = 2005*24
5*401*2*2*2*5 = 2005*60

this could go on forever. Ther are infinites N's where this would work.

2005*16 has too many divisors, namely 1,2,4,8,10,16,20,...

Try M = 5: Then 1, 5, 25, 401, 2005, 10025 are the only divisors of 2005*5.
M = 401 also works.
If M has any prime factor other than 5 or 401, let's say p, there will be at least seven divisors: 1, 5, 401, p, 5p, 401p, 2005p.

If M is a power of 5 or has divisors 5 and 401, there will also be too many divisors.

Thanks for the help

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