20190121, 10:15  #1 
Mar 2018
17·31 Posts 
2^x41
Are 8=x and 36=x the only integer values such that 2^x41 is divisible by 215?

20190121, 10:36  #2  
Sep 2002
Database er0rr
2×11×167 Posts 
Quote:
Code:
n=215;ph=eulerphi(n);e=8+ph*2^1000;r=Mod(2,n)^(e)41;if(r==0,print(e)) *** _^_: Warning: Mod(a,b)^n with n >> b : wasteful. 1800134460072929099193354082420803041743160083665296460505500652462189765889892685788573276410369041654359050501489286666314404079963087593245409032549360567061007386170470805550740533650478351835496302601711807797241766132922992949939811382940946569647961008251339084166952211717640896988650552235655176 For those values less than or equal to eulerphi(215): Code:
n=215;ph=eulerphi(n);for(k=1,ph,r=Mod(2,n)^(k)41;if(r==0,print(k))) 8 36 64 92 120 148 Last fiddled with by paulunderwood on 20190121 at 10:55 
