April 11th, 2014, 01:32 AM  #1 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  Reminder
Reminder of 3^50 divide by 7

April 11th, 2014, 01:48 AM  #2 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 
Hint: $\displaystyle 3^6\equiv1\!\pmod7$ (Fermatâ€™s little theorem).

April 11th, 2014, 02:04 AM  #3 
Senior Member Joined: Apr 2014 From: UK Posts: 918 Thanks: 331 
I make the remainder 2

April 11th, 2014, 03:04 AM  #4 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
You can also use $$3^3 \equiv 1 \ (mod \ 7) $$ $$(3^3)^{16} \equiv (1)^{16} \ (mod \ 7)$$ $$3^{48} \equiv 1 \ (mod \ 7)$$ $$3^2 \cdot 3^{48} \equiv 3^2 \cdot 1 \ (mod \ 7)$$ $$3^{50} \equiv 9 \ (mod \ 7)$$ $$3^{50} \equiv 2 \ (mod \ 7)$$ Alternatively , $$3^{1} \ (mod \ 7) \ = \ 5$$ Because $$3 \cdot 5 \equiv 1 \ (mod \ 7)$$ $$3^3 \equiv 1 \ (mod \ 7)$$ $$3^{51} \equiv 1 \ (mod \ 7)$$ $$3^{1} \cdot 3^{51} \equiv 3^{1} \cdot (1) \ (mod \ 7)$$ $$3^{50} \equiv 5 \cdot (1) \ (mod \ 7)$$ $$3^{50} \equiv 5 \ (mod \ 7)$$ $$3^{50} \equiv 2 \ (mod \ 7)$$ I left out a few minor details which I'm sure you can work out yourself , if not , please feel free to ask for clarification. The trick is to get congruent to $ \pm 1$ , then use IF $$a \equiv b \ (mod \ m)$$ THEN $$a^n \equiv b^n \ (mod \ m) $$ 

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