How to know if a term is divisible by 10 I have some difficulties to solve this easy problem, could someone help me?
Is $4^{1000}-6^{500}$ divisible by $10$?
 A: we have $4^{1000}-6^{500}\equiv 6-6=0\equiv 0 \mod 10$
A: We can see that $4^1 = 4$, $4^2 = 16$, $4^3 = 64$, etc. It starts out with 4 as the last digit, which when multiplied by 4 gives you a 6, which when multiplied by a 4 gives you a 4, setting up the pattern of 6 as the last digit for even powers. Therefore $4^{1000}$ has 6 as its last digit. Clearly all powers of 6 have 6 as their last digit because if you multiply something that ends with 6 by 6 the product will end with a 6 since $6\times6=36$. Since the last digits are both 6, when you subtract you get 0 as the last digit, which means the difference is divisible by both 5 and 2 and is therefore divisible by 10.
A: $4^2 \equiv 6 \mod 10$
$\Rightarrow 4^{1000} \equiv 6^{500} \mod 10$
A: Your expression can be written as $16^{500}-6^{500}$. Recall that $a^n-b^n$ is divisible by $a-b$ for $n \in \mathbb{N}$.
A: HINT: What is the right-most digit of $4^{1000}$? What is the right-most digit of $6^{500}$? What does that tell you about the right-most digit of $4^{1000}-6^{500}$?
