Question

find the unit digit of
$\rm 27^{2015} + 21^{2015} -23^{2015}$
​

Answer 1

$\Huge\textrm{7}$

$\;$

$\Large\textrm{Explanation}$

$\rm\dots7^{0}=\dfrac{\rm\dots7}{\rm\dots7}\ \ \ =\ \ \ \ 1$

$\rm\dots7^1=\dots7\cdot1=\dots7$

$\dots7^2=\dots7\cdot7=\dots9$

$\dots7^3=\dots9\cdot7=\dots3$

The cycle starts over every 4th number.

$\;$

$\rm\dots1^{0}=\dfrac{\rm\dots1}{\rm\dots1}\ \ \ =\ \ \ \ 1$

The number with 1 in the ending digit always ends in 1.

$\;$

$\rm\dots3^{0}=\dfrac{\rm\dots3}{\rm\dots3}\ \ \ =\ \ \ \ 1$

$\rm\dots3^1=\dots3\cdot1=\dots3$

$\dots3^2=\dots3\cdot3=\dots9$

$\dots3^3=\dots3\cdot9=\dots7$

The cycle starts over every 4th number.

$\;$

$\rm2015=4\cdot503+3$

$\rm27^{3}$ ends with a $\rm3$.

$\;$

$\rm2015=1\cdot2015+0$

$21^{0}$ ends with a $\rm0$.

$\;$

$\rm2015=4\cdot503+3$

$23^{3}$ ends with a $\rm7$.

$\;$

$27^{2015}+21^{2015}-23^{2015}$

The number is big enough for subtraction.

$\;$

Hence,

$\rm\dots3+\dots1-\dots7$

$\;$

$\rm=\dots3+\dots1+(\dots10-\dots7)$

$\;$

$\rm=\dots3+\dots1+\dots3$

$\;$

$\rm=7$

Answer 2

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