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 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 Thanks from skeeter
 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
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Quote:
 Originally Posted by mathman $\displaystyle 2^5\equiv 2\pmod{10}$ $\displaystyle 2009\equiv 1 \pmod 4$
I should have said:
$\displaystyle 2^{4n+1}\equiv 2\pmod{10}$

 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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