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Which would be the last digit for $3^{2019}$ ?
You can
check last digits for $3^x$ with $x=\{0,1,2,3,4,5,6\}$ and see if something is repeated.
And afterwards
think about modulus for the exponent number and the pattern found.
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6
Because
$3^4=81\equiv1 \:(\text{mod}\;10)$,
and
$2019=(4\times504)+3\equiv3\:(\text{mod}\;4)$,
we have
$3^{2019}=(3^4)^{504}\times3^3\equiv(1)^{504}\times3^3=27\equiv7\:(\text{mod}\;10)$,
so the answer is
$7$.
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1
As we know,
Powers of 3 are numbers ending in $\{1,3,9,7\}$ sequentially.
As
$MOD(2019,4)=3$
So we know that
The last digit for $3^{2019}$ will be the forth in the sequence stated in the beginning. As if result was 0 it would have been the first element.
So the result is that the last digit for $3^{2019}$ is:
$7$
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3
The list of last digits:
$3$
$9$ ($3×3$)
$7$ ($3×3×3$)
$1$ ($3×3×3×3$)
And then the cycle repeats again.
So, 3 to the 2019 th power is same as $3^3$(There is a cycle of 4).
Which gives 27,which have a last digit of 7.
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