<s>[INST] Find the unit digit of $2124^{392}+3143^{394}*7177^{392}-8818^{394}$ What I do here is I find the remainder of each parcel in mod 10 and then do the maths and mod 10 again:
$2124^{392} \equiv 4^{392} \pmod{10}$
$$4^1 \equiv 4 \pmod{10} \\
4^2 \equiv 6  \\
4^3 \equiv 4 \\
4^4 \equiv 6 \\
(...)$$
Then I do $392 \pmod{2}$. Problem is that here, it's zero. If it were 1 or 2, the remainder would be either 4 or 6. How do I solve it this way?
 [/INST] Using Newton's  binomial formula,
$$4^{392}=(10-6)^{392} $$
$$=6^{392} \mod 10 =6 \mod 10$$
Since $$6^2 = 6\mod 10$$
For the second and the third , $ 3$ and $ 7$ are coprime with $ 10$ so by Euler's  Theorem 
$$3^{\phi(10)}=1 \mod(10) =3^4$$
and
$$7^4 = 1 \mod 10$$
For example
$$3^{394}=3^{98.4+2}=9 \mod 10$$
For the last one, observe that
$$2^5 = 2 \mod 10$$
then
$$8^5= 2^{3.5}=2^3=8 \mod 10$$
thus
$$8^{394}=8^{78.5+4}=8^{78+4}=8^{16.5+2}=8^{16+2}=8^{3.5+3}=8^6$$
$$=8^{5+1}=8^2=4 \mod 10$$
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