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Neighbors (
Posted on 20140418
)
Find the smallest integer
n
such that both
n1
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 n1 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 20140418 10:29:22
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