My Calculator Can't Handle This Monster

Quite easy by modular arithmetic,

$23^2\equiv-1 (mod 10)$

$23^6\equiv(-1)^3\equiv-1\equiv9 (mod 10)$

And we have unit digit of $6^{23}$ be 6.

Unit Digit : $9 + 6 = 15$ i.e $5$

$\begin{aligned} 23^6\equiv 3^6 & \equiv 9\pmod{10}\\ 6^{23} & \equiv 6 \pmod{10}\\ \implies 23^6+6^{23} & \equiv 5 \pmod{10} \end{aligned}$

To find unit digit of 23^6 + 6^23

To find unit digit of a^x, here a is base and x is index, always find the remainder of x divided by 4,

If Remainder is 1 then unit digit is (unit digit of base)^1

If Remainder is 2 then unit digit is (unit digit of base)^2

If Remainder is 3 then unit digit is (unit digit of base)^3

If Remainder is 0 then unit digit is (unit digit of base)^4

In case of 23^6 here 6/4 remainder is 2 hence unit digit of unit digit is 3^2 = 9

In case of 6^23 here 23/4 remainder is 3 hence unit digit of unit digit is 6^3 = 6

So answer is 9 + 6 = 15 and unit digit is 5

Note unit digit of 0, 1, 5, 6, raised to any power unit digit remains same respectively.

Now why index/4 ?, take any digit say 2,

unit digit of 2^1 = 2, 2^2 = 4, 2^3 =8, 2^4 = 6, 2^5 = 2, ...

see unit digit is 2, 4, 8, 6, 2, ... repeats.

The trick is.. 1st you need to find a pattern of the units place for successive powers of any number. In this case 3 and 6.
if we look at 3;
3^1=3 , 3^2=9, 3^3=27, 3^4=81, 3^5=243.....
So the pattern is 3,9,7,1
Hence we write 3^4 ×3^2
this gives units place as 1×9=9
For 6 it always ends in 6 for any power.
Hence 6^23=6 as units place
just add the units place digits
I.e 9+6=15
thus the answer 5.

→23^6 = 3^6 = 9
→6^23 = 6
→9+6=15
→ 5