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May 7th, 2015, 02:14 PM   #1
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Last digit

find the last digit of
2^2009
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May 7th, 2015, 02:53 PM   #2
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May 7th, 2015, 05:43 PM   #3
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$\displaystyle 2^5\equiv 2\pmod{10}$

$\displaystyle 2009\equiv 1 \pmod 4 $
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May 8th, 2015, 04:40 AM   #4
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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?
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May 8th, 2015, 01:30 PM   #5
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Quote:
Originally Posted by mathman View Post
$\displaystyle 2^5\equiv 2\pmod{10}$

$\displaystyle 2009\equiv 1 \pmod 4 $
I should have said:
$\displaystyle 2^{4n+1}\equiv 2\pmod{10}$
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May 8th, 2015, 06:58 PM   #6
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As $2^{2008} = 16^{502}$ has last digit 6, $2^{2009}$ has last digit 2.
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