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A Divisor Diversion (Posted on 2019-03-27) Difficulty: 3 of 5
Find the sum of all positive integers n, each of which has precisely 16 positive divisors and a successor n+1 that has precisely 3 divisors.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 5.0000 (1 votes)

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solution | Comment 3 of 5 |
n=p^2 - 1 = (p+1)(p-1).

n+1 will be the square of a prime.  Write n=p^2 - 1. 

If p=6a+1, n=(2^2)*(3)*(a)*(3a+1).  The factor (2^2) means a factor of 3 in the total divisors so another factor of 2 is needed. That would get the contribution of (2^3)*(3) to 8 towards a goal of 16, leaving a needed factor of 2.  But unless a=2, (a) and (3a+1) each contribute a factor of at least 2.  So a=2 and n=168.

Similarly, if p=6a-1, n=(2^2)*(3)*(a)*(3a-1), an additional factor of 2 is needed, and a total of 16 divisors is possible only if a=2, giving n=120.

As p=3 doesn't work there are no more solutions and the sum=288.  


 

  Posted by xdog on 2019-03-28 17:57:25
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