May 7th, 2015, 02:14 PM  #1 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  Last digit
find the last digit of 2^2009 
May 7th, 2015, 02:53 PM  #2 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361 
2

May 7th, 2015, 05:43 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,607 Thanks: 616 
$\displaystyle 2^5\equiv 2\pmod{10}$ $\displaystyle 2009\equiv 1 \pmod 4 $ 
May 8th, 2015, 04:40 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Did you try at all? An obvious first step would be to look at powers of 2 to see there were any pattern. 2^1= 2: last digit 2 2^2= 4: last digit 4 2^3= 8: last digit 8 2^4= 16: last digit 6 2^5= 32: last digit 2 2^6= 64: last digit 4 2^7= 128: last digit 8 2^8= 256: last digit 6 2^9= 512: last digit 2 2^10= 1024: last digit 4 2^11= 2048: last digit 8 2^12= 4096: last digit 6 Does that give you any ideas? 
May 8th, 2015, 01:30 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,607 Thanks: 616  
May 8th, 2015, 06:58 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,698 Thanks: 1804 
As $2^{2008} = 16^{502}$ has last digit 6, $2^{2009}$ has last digit 2.


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