20200930, 13:44  #1 
Mar 2018
1022_{8} 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?

20200930, 14:17  #2  
"Robert Gerbicz"
Oct 2005
Hungary
2·3^{2}·83 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. 

20201010, 15:38  #3 
"Ruben"
Oct 2020
Nederland
46_{8} 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  20151117 09:22 
I think I found proof that no odd perfect numbers exist!  Philly314  Aliquot Sequences  3  20141116 14:58 
Proof of Primality Test for Fermat Numbers  princeps  Math  15  20120402 21:49 
PRIMALITY PROOF for Wagstaff numbers!  AntonVrba  Math  96  20090225 10:37 
Proof of Perfect Numbers  Vijay  Miscellaneous Math  10  20050528 18:11 