Task B-1.1. Determine the remainder when the number $\left(7^{2012}\right)^{2014}-\left(3^{12}\right)^{14}$ is divided by 10.
Solution.

$$
7^{1}=7,7^{2}=49,7^{3}=343,7^{4}=2401,7^{5}=16807, \ldots
$$

Therefore, the powers of the number 7 end with the digits $7,9,3,1$.

$$
3^{1}=3,3^{2}=9,3^{3}=27,3^{4}=81,3^{5}=243, \ldots
$$

The powers of the number 3 end with the digits $3,9,7,1$.

The number 2012 is divisible by 4, so the number $7^{2012 \cdot 2014}$ has the last digit the same as $7^{4}$, which is the digit 1.

The number 12 is divisible by 4, so the number $3^{12 \cdot 14}$ has the last digit the same as $3^{4}$, which is the digit 1.

If we subtract the numbers $7^{2012 \cdot 2014}$ and $3^{12 \cdot 14}$, the last digit of the difference will be 0, so the given number is divisible by 10.