All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Purely prime (Posted on 2013-02-19)
Twenty-one prime numbers are in arithmetic sequence with difference d. Prove that d is divisible by 9699690.

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Primes Comment 3 of 3 |
Let p1, p2, ..., p21 be the prime numbers. Then, p1+d=p2, p2+d=p3, ..., p20+d=p21. Take any prime p<10. If d is not divisible by p, then at least 2 of the prime numbers are divisible by p. Since p is the only prime divisible by p, that is impossible. Therefore, d is divisible by p. Then, d is divisible by 2*3*5*7=210. Take any prime 10<p<20. If d is not divisible by p, then at least 1 of the prime numbers is divisible by p. Since p is the only prime divisible by p, pn=p for some n. Since d is divisible by 210, p2>210, ..., p21>210. Since p<20, p2>p, ..., p21>p. Therefore, p1=p. Then, p1+pd is one of the primes. That is impossible. Therefore, d is divisible by p. Then, d is divisible by 2*3*5*7*11*13*17*19=9699690.

 Posted by Math Man on 2016-09-05 21:00:42
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 (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information