All about
flooble

fun stuff

Get a free chatterbox

Free JavaScript

Avatars
perplexus
dot
info
Home
>
Numbers
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
No Rating
Comments: (
Back to comment list
 You must be logged in to post comments.
)
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
Please log in:
Login:
Password:
Remember me:
Sign up!

Forgot password
Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ

About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
blackjack
flooble's webmaster puzzle
Copyright © 2002  2017 by
Animus Pactum Consulting
. All rights reserved.
Privacy Information