Hoe to solve this?We have #hat3 inZZ_7#;Calculate #hat3^6#and#hat3^2014#

1 Answer
George C.
Apr 6, 2017

#hat(3)^6 = hat(1)#

#hat(3)^2014 = hat(4)#

Explanation:

Writing #hat(a)# for the equivalence class of #a# modulo #7#, we find:

#hat(3)^0 = hat(1)#

#hat(3)^1 = hat(3)#

#hat(3)^2 = hat(3*3) = hat(9) = hat(2)#

#hat(3)^3 = hat(3*2) = hat(6)#

#hat(3)^4 = hat(3*6) = hat(18) = hat(4)#

#hat(3)^5 = hat(3*4) = hat(12) = hat(5)#

#hat(3)^6 = hat(3*5) = hat(15) = hat(1)#

We could have deduced #hat(3)^6 = hat(1)# from Fermat's little theorem, which can be stated:

If #p# is a prime number, then for any integer #a#:

#a^p -= a" "# modulo #p#

If in addition #a != 1# modulo #p#, then we can divide by #a# to find:

#a^(p-1) -= 1" "# modulo #p#

From #hat(3)^6 = hat(1)#, we can deduce:

#hat(3)^(6m+n) = hat(3)^n#

So:

#hat(3)^2014 = hat(3)^(6*335+4) = hat(3)^4 = hat(4)#

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