|
2020-09-30, 13:44
|
#1 |
|
Mar 2018
72×11 Posts |
How to proof that numbers 18, 108, 1008,...will never be divisible by 6^4?
How to proof that numbers of the form 18, 108, 1008, 10008, 100008, 1000...0008 will never be divisible by 6^4?
|
|
|
|
2020-09-30, 14:17
|
#2 | |
"Robert Gerbicz"
Oct 2005
Hungary
32×179 Posts |
Quote:
a(n)=10^n+8==8 mod 16 hence it won't be divisible by even 16=2^4 so not by 6^4. And you can check the n<=3 cases easily since 6^4=1296>1008. |
|
|
|
|
2020-10-10, 15:38
|
#3 |
|
"Ruben"
Oct 2020
Nederland
2×19 Posts |
6^4
Or another way is that numbers ending in 1, 5 and 6 multiplied by a number ending by their end digit ends in that digit!
|
|
|
|
 |
| Thread Tools | |
Similar Threads
|
||||
| Thread | Thread Starter | Forum | Replies | Last Post |
| Article: First proof that infinitely many prime numbers come in pairs | Paulie | Twin Prime Search | 46 | 2015-11-17 09:22 |
| I think I found proof that no odd perfect numbers exist! | Philly314 | Aliquot Sequences | 3 | 2014-11-16 14:58 |
| Proof of Primality Test for Fermat Numbers | princeps | Math | 15 | 2012-04-02 21:49 |
| PRIMALITY PROOF for Wagstaff numbers! | AntonVrba | Math | 96 | 2009-02-25 10:37 |
| Proof of Perfect Numbers | Vijay | Miscellaneous Math | 10 | 2005-05-28 18:11 |
