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 April 11th, 2014, 01:32 AM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 Reminder Reminder of 3^50 divide by 7 Thanks from agentredlum
 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: $\displaystyle 3^6\equiv1\!\pmod7$ (Fermat’s little theorem). Thanks from agentredlum
 April 11th, 2014, 02:04 AM #3 Senior Member   Joined: Apr 2014 From: UK Posts: 918 Thanks: 331 I make the remainder 2 Thanks from agentredlum
 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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