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