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Neighbors (Posted on 2014-04-18) Difficulty: 2 of 5
Find the smallest integer n such that both n-1 and n+1 have the same number of divisors as n.
Bonus task: List few additional numbers with said feature.

See The Solution Submitted by Ady TZIDON    
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solution | Comment 1 of 8
Since d(p)=2 for prime p, none of the series can be prime, so I looked for n=2p where n-1 and n+1 are of the form pq, a product of primes.

A little work with pen and paper finds 33,34,35, all with number of divisors=4.

  Posted by xdog on 2014-04-18 10:29:22
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