Finding remainders using modulo

Question

Determine the remainder of $2014^{2015} \cdot 2016^{2017} + 2018^{2019}$ divided by 13.

I can't figure out how to manipulate the 2018 part to get it to some form of 13. Any suggestions?

Answer 1

We have: $2014^{2015}\cdot 2016^{2017} = 12^{2015}\cdot 1^{2017} (\mod 13) = 12\cdot 1(\mod 13) = 12(\mod 13)$, and $2018^{2019} = 3^{2019} = (3^{12})^{168}\cdot 3^3 = 1^{168}\cdot 1 = 1(\mod 13)$. Adding these mods we have: $S = (12+1)(\mod 13) = 13(\mod 13) = 0(\mod 13)$.

Answer 2

2014 is $(-1)$ modulo 13. 2016 is 1 modulo 13. 2018 is 3 modulo 13, but then $2018^3$ is $3^3$ which is 27 which is 1 modulo 13. After this, you can work with the powers very easily, just remember to take the third power of 2018 into account.

Answer 3

As $2014\equiv-1\pmod{13},(2014,13)=1$ and as $2015\equiv11\pmod{\phi(13)}$

Using Fermat's Little Theorem, $2013^{12}\equiv1\pmod{13}$

$\implies2013^{2015}\equiv(-1)^{11}\equiv-1\pmod{13}$

Similarly, $2018^{2019}\equiv3^3\pmod{13}\equiv1$ and $2016^{2017}\equiv1^1$