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.
A number can only have 3 factors if it is the square of a prime: n+1=t^2 for prime t.
A number with 16 factors must have the following prime factorizations: p^15, q*p^8, q^3*p^3, r*q*p^4, s*r*q*p
n=t^2-1=(t+1)(t-1)
After ruling out t=2 we see that n has at least 2^3 and no other prime factor can appear more than once.
So n is either 2^15 or of the form q*2^8, r*q*2^4, s*r*q*2.
2^15 doesn't work. I don't have time to check the other cases.
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Posted by Jer
on 2019-03-27 14:56:21 |
No Subject
