My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum


Reply
 
LinkBack Thread Tools Display Modes
April 2nd, 2013, 02:23 AM   #1
Member
 
Joined: Apr 2013

Posts: 65
Thanks: 0

A Modular Arithmetic Question

What are the last 2 digits of when written in base 3?
Drake is offline  
 
April 2nd, 2013, 06:25 AM   #2
Member
 
Joined: Mar 2013

Posts: 90
Thanks: 0

Re: A modular-arithmetic question

and so . Hence , say .

Now and .

Hence , i.e. the last two digits in base 3 are 21.
Nehushtan is offline  
April 2nd, 2013, 06:45 AM   #3
Member
 
Joined: Apr 2013

Posts: 65
Thanks: 0

Re: A Modular Arithmetic Question

Oh I see. Thanks a lot, I tried to use mod 100 and then turn the result into the third base but it clearly didn't work so well. Base conversions always confuse me.
I have a question though, shouldn't we use instead mod 8 instead of 9? What I mean by that is:
and
Drake is offline  
April 2nd, 2013, 07:10 AM   #4
Member
 
Joined: Mar 2013

Posts: 90
Thanks: 0

Re: A modular-arithmetic question

We take mod 9 because weíre working in base 3 and (just as if you wanted the last two digits in base 10 you would take mod 100 as ).

By the way, my original proof was incorrect in some details. Iíve edited and fixed my post.
Nehushtan is offline  
April 2nd, 2013, 07:33 AM   #5
Member
 
Joined: Apr 2013

Posts: 65
Thanks: 0

Re: A Modular Arithmetic Question

Yeah now it's clear, thanks. I meant btw, a little brain malfunction there. So we can either take mod 3 or 6 because both and , thats what I tried to mean.

I have one last question. Last 2 digits of ?
Drake is offline  
April 2nd, 2013, 11:57 AM   #6
Member
 
Joined: Mar 2013

Posts: 90
Thanks: 0

Re: A modular-arithmetic question

Well, so try and work around that.
Nehushtan is offline  
April 2nd, 2013, 12:19 PM   #7
Global Moderator
 
CRGreathouse's Avatar
 
Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 933

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Re: A modular-arithmetic question

Quote:
Originally Posted by Nehushtan
By the way, my original proof was incorrect in some details. Iíve edited and fixed my post.
I think it's still wrong -- or else I'm making a mistake. n^6 = 1 mod 9 for n not divisible by 3, so 2005^(2003^2004) = 2005^(2003^2004 mod 6) mod 9. 2003^2004 is 1 mod 6, so 2005^(2003^2004) = 2005 = 7 mod 9. Adding 3 you get 1, so the last two digits are 01.
CRGreathouse is offline  
April 2nd, 2013, 12:33 PM   #8
Member
 
Joined: Apr 2013

Posts: 65
Thanks: 0

Re: A Modular Arithmetic Question

CRGreathouse: I guess you missaw it, it should be 2005^(2003^(2004)+3) instead of 2005^(2003^2004) + 3. So 2003^2004 is 1 mod 6, adding 3 you get 4. 2005^(6k+4)=7 (mod 9)

Nehushtan: How do you know that 9^10=1 (mod 100) ?
Drake is offline  
April 2nd, 2013, 12:55 PM   #9
Senior Member
 
Joined: Mar 2012

Posts: 571
Thanks: 26

Re: A Modular Arithmetic Question

Quote:
Originally Posted by Drake
Nehushtan: How do you know that 9^10=1 (mod 100) ?
Have a look at the pattern (mod 100) of 9^1, 9^2, 9^3 - you should spot it fairly early on.
Hedge is offline  
April 2nd, 2013, 01:11 PM   #10
Member
 
Joined: Apr 2013

Posts: 65
Thanks: 0

Re: A Modular Arithmetic Question

Alright thanks. I tried to take mod 40 but it didn't work out.
Drake is offline  
Reply

  My Math Forum > College Math Forum > Number Theory

Tags
arithmetic, modular, question



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Modular Arithmetic aaron-math Number Theory 3 August 27th, 2012 08:51 AM
Modular arithmetic mathslog Number Theory 1 April 6th, 2012 03:08 PM
Question on Modular Arithmetic Hanny David Number Theory 0 January 24th, 2009 03:48 PM
Modular arithmetic help Ciachyou Number Theory 1 January 23rd, 2009 11:58 AM
A Modular Arithmetic Question Drake Algebra 4 December 31st, 1969 04:00 PM





Copyright © 2017 My Math Forum. All rights reserved.