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The digit in the unit's place of the number $2017^{2017}$ is

  1. $1$
  2. $3$
  3. $7$
  4. $9$
in Numerical Ability by Veteran (105k points)
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2 Answers

+1 vote
7 cyclicity =4

2017/4 remainder =1

(7)^1 = 7

Answer= 7
by (431 points)
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To get the unit digit of $2017^{2017}$, we will find $2017^{2017}\;mod\;10.$

$2017^{2017}\;mod\;10$$=2017^{2016+1}\;mod\;10$

$=(2017^{2016}*2017^{1})\;mod\;10$

$=(2017^{2016}\;mod\;10*2017^{1}\;mod\;10)\;mod\;10$

$=((2017^{4})^{404}\;mod\;10*2017^{1}\;mod\;10)\;mod\;10$

$=(1*2017^{1}\;mod\;10)\;mod\;10$    ( $\because$ when $a,n$ are co-primes then $a^{\Phi (n)}mod\;n=1$. Here $2017,10$ are co-primes and $\Phi (10)=4$. So, $2017^{4}\;mod\;10=1$ ) $[\Phi(n) =  Euler \;\;Totient\;\; Function ]$

$=(2017\;mod\;10)\;mod 10$

$=7\;mod\;10$

$= 7$

option c is the answer..
by Active (4.6k points)

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