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 Algebra Pre-Algebra and Basic Algebra Math Forum

 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: 963 Thanks: 342 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 $$a^n \equiv b^n \ (mod \ m)$$  Tags reminder Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post johnny New Users 7 August 17th, 2007 04:54 PM

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