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.
Obviously, n+1 is the square of a prime. Suppose n+1=p^2. Then, n=p^2-1. If p=2, then n=2^2-1=4-1=3. Then, n has 2 divisors. If p=3, then n=3^2-1=9-1=8. Then, n has 4 divisors. Therefore, p>=5.
Since p is prime and p>=5, then p+1 and p-1 are both even. Then, either p+1 or p-1 is divisible by 4. Also, either p+1 or p-1 is divisible by 3. Therefore, n=p^2-1=(p+1)(p-1) is divisible by 2*3*4=24. Let n=24x.
The only way that n can have 16 divisors is if x=9, 16, or a prime greater than 3. If x=9, then n=216. Then, n+1=217 is not a square. If x=16, then n=384. Then, n+1=385 is not a square. Therefore, x is a prime greater than 3.
If p=5, then n=24, so x=1. If p=7, then n=48, so x=2. If p=11, then n=120, so x=5. If p=13, then n=168, so x=7. If p>13, then p+1 and p-1 both have an extra factor other than 2, 3, and 4. Then, x is composite. Therefore, p=11 or 13. Then, n=120 or 168. The sum of all n is 120+168=288.
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Posted by Math Man
on 2019-03-29 17:02:38 |
Answer
