Find the last digit of $7^{802} -3^{683}$ 
Find the last digit of $$7^{802} -3^{683}$$

$$7^{802} -3^{683}=(50 -1)^{401} -3(10 -1)^{341}$$
$$=50k -1 -3(10m -1) \space \text{where k,m are fixed integer } $$
$$=50k -30m +2 $$
So,the last digit will be $2$
Someone just added this question, but deleted it. I solved this question and got this thing. I just want to know if my answer correct or not.

Okay, after reading the comments I realize my answer is completely out of the experience level of the OP.  But read ahead for a taste of exciting things to come.
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Very nice!
I want to introduce you to the idea of Fermat's Little Thereom however.
If $\gcd(a, m) = 1$.  And $\phi(m) = $ the number of natural numbers less than $m$ that are relatively prime to $m$ then:
$a^{\phi (m)} \equiv 1 \mod m$
(or in other words $a^{\phi(m)} = km + 1$ for some $k$.)
So... as $1,3,7$ and $9$ are relatively prime to $10$ so $\phi(10) = 4$.  And $\gcd(7,10) = \gcd(3,10) = 1$.
So $7^4 = 1 \mod 10$ ($7^4 = 2401$) and $3^4 \equiv 1 \mod 10$ ($3^4 = 81$).
So $7^{802} - 3^{683} = 7^{2 + 4k} - 3^{3 + 4j}$
$= 7^2*(7^4)^k - 3^3*(3^4)^j \equiv 7^2*1^k - 3^3*1^j \mod 10$
$\equiv 49 - 27 \mod 10 \equiv 9 -7 \mod 10 = 2 \mod 10$
So $2$ is the last digit.
.........
Okay, I admit, now that I've typed the whole thing out that seems a LOT more complicated than your elegant fast solution.
But it's useful to know.
If I asked "what is the remainder when $8^{386} - 6^{216}$ is divided by $13$" it might be easier to use Fermat's Little Theorem, rather than "$8^2 = 64 = 5*13 - 1$ and $6^6 =  46656 = 3589*13 - 1$ so $(-1)^{193} - (-1)^36 = -1 -1 =-2$ so $11$ is the remainder"
[$13$ is prime so all numbers less than $13$ are relatively prime so $\phi(13) = 2$.  So $8^{386} \equiv 1 \mod 13$ and $6^216 \equiv 1 \mod 13$.
[So $8^{386=2 + 12k} - 6^{216=12j} \equiv 8^2 - 1 \mod 13$
[$\equiv 63 \mod 13 \equiv 11 \mod 13$]
