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 Find the units digit of 3^2019 in base 16.

 May 13, 2022
 #1
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3^2019 mod 10^10==4B5998A3F1C1AAA3 - these are the 10 "digits" in base 16

 May 14, 2022
 #2
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\(3^0=1 \;\;(base10)\;\;\equiv 1 \;\;(base16)\;\; \\ 3^1=3 \;\;(base10)\;\;\equiv 3 \;\;(base16)\;\; \\ 3^2= \;\;(base10)\;\;\equiv 9 \;\;(base16)\;\; \\ 3^3=27 \;\;(base10)\;\;\equiv B \;\;(base16)\;\; \\ 3^4=81 \;\;(base10)\;\;\equiv 51 \;\;(base16)\;\; \\ \)

 

The pattern for the last digit is now set.

 

\(3^{0+4n} \;\;ends \;\;in\;\; 1 \;\;(base16)\;\; \\ 3^{1+4n} \;\;ends \;\;in\;\; 3 \;\;(base16)\;\; \\ 3^{2+4n}\;\;\;\;ends \;\;in\;\; 9 \;\;(base16)\;\; \\ 3^{3+4n} \;\;\;\;ends \;\;in\;\;B \;\;(base16)\;\; \\ \)

 

2019 = 4*504+3

So 

\(3^{2019}\quad \text{has a unit digit of WHAT in base 16.}\)

 

 

** Please give a response

 May 14, 2022
edited by Melody  May 14, 2022

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