A apple2010 Mar 2010 8 0 May 12, 2010 #1 Give two different methods for finding 2^999(mod 5), one using Fermat's little theorem and the other using basic problem solving methods.

Give two different methods for finding 2^999(mod 5), one using Fermat's little theorem and the other using basic problem solving methods.

roninpro Nov 2009 485 184 May 12, 2010 #2 I'll give you a hint for a basic method. Write \(\displaystyle 2^{999}\equiv 2\cdot 2^{998}\equiv 2\cdot 4^{499}\pmod{5}\). Can you compute this directly?

I'll give you a hint for a basic method. Write \(\displaystyle 2^{999}\equiv 2\cdot 2^{998}\equiv 2\cdot 4^{499}\pmod{5}\). Can you compute this directly?

F firebio Jul 2009 69 6 May 12, 2010 #3 apple2010 said: Give two different methods for finding 2^999(mod 5), one using Fermat's little theorem and the other using basic problem solving methods. Click to expand... Using Fermats little theorem \(\displaystyle 2^4 \equiv 1 \pmod{5} \) Then \(\displaystyle (2^4)^{249} \equiv 1 \pmod{5} \) Finally \(\displaystyle (2^4)^{249}*2^3 \equiv 1*2^3 \pmod{5} \) therefore \(\displaystyle 2^3 \equiv 3 \pmod{5} \) So \(\displaystyle 2^{999} \equiv 3 \pmod{5} \)

apple2010 said: Give two different methods for finding 2^999(mod 5), one using Fermat's little theorem and the other using basic problem solving methods. Click to expand... Using Fermats little theorem \(\displaystyle 2^4 \equiv 1 \pmod{5} \) Then \(\displaystyle (2^4)^{249} \equiv 1 \pmod{5} \) Finally \(\displaystyle (2^4)^{249}*2^3 \equiv 1*2^3 \pmod{5} \) therefore \(\displaystyle 2^3 \equiv 3 \pmod{5} \) So \(\displaystyle 2^{999} \equiv 3 \pmod{5} \)