Can you guess the unit’s digit?

There is a simple way to find the last digit of $a^b$ , where the last digit of $a$ is $x$ .

Last digit of $a^b =$ last digit of $x^{b \bmod {4}}$

Note: If the ${b \bmod 4} = 0$ , then let $b = 4$ .

Ex: Last digit of $17^{17} = 7^{17} = 7^{17 \bmod {4}} = 7^1 = 7$

Ex: Last digit of $973^{2008} = 3^{2008} = 3^{2008 \bmod {4}} = 3^4 = 81 = 1$

Now let's apply this to the problem:

$3^{146} = 3^{146 \bmod {4}} = 3^2 = \color{#D61F06}{9}$

$5(14^{39}) = 5(4^{39}) = 5(4^{39 \bmod {4}}) = 5(4^3) = 5 \times 64 = 5 \times 4 = 2\color{#D61F06}{0}$

$6^{48} = 6^{48 \bmod {4}} = 6^4 = 129\color{#D61F06}{6}$

$2^{2005} = 2^{2005 \bmod {4}} = 2^1 = \color{#D61F06}{2}$

Thus, the units digit of $3^{146} + 5(14^{39}) - 6^{48} + 2^{2005}$ =

$9 + 0 - 6 + 2 = 5$