Please help:

Find all composite integersMsuch that the product 2005 *Mhas exactly eight divisors.

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- April 29th 2005, 02:26 PM #1

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- April 29th 2005, 07:54 PM #2Originally Posted by
**mathsmad**

If M is allowed to be prime, any prime number not equal to 5 or 401 will do.

If M is composite and has a prime factor other than 5 or 401, there will be too many factors. Similar if M is divisible by 5*5*401 or 5*401*401.

So only powers of 5 or 401 remain. Counting divisors for the cases M = 5^k, k > 2 and M = 401^k, k > 2 shows that these have too many divisors. This implies that the cases given above are the solution set.

- May 14th 2005, 08:48 PM #3

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