Modular Arithmetic Involving Remainders

Question

Using modular arithmetic, solve the following:

Find the remainder of $(2014^{2015} \cdot 2016^{2017}) + 2018^{2019}$ when it is divided by 13.

I have no idea where to start. I've tried putting this in (mod 13) but I'm not sure where to start or even finish.

Any help would be appreciated.

ajotatxe · Answer 1 · 2014-10-28T23:32:11.687

1

You need the little Fermat's theorem (or very much patience...). This theorem guarantee that, in particular, $x^{13}\equiv x\pmod {13}$. Or more generally, $$x^{n+12k}\equiv x^n\pmod {13}$$ for any integer $k$.

Now, $2014\equiv -1\pmod{13}$, $2016\equiv 1\pmod {13}$ and $2018\equiv 3\pmod {13}$, so

$$2104^{2015}\cdot2016^{2017}+2018^{2019}\equiv -1\cdot 1+3^{168\cdot12+3}\equiv -1+3^3\equiv 0\pmod{13}$$

edited Oct 28 '14 at 23:32

answered Oct 28 '14 at 19:15

ajotatxe

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How is -1 + 3^3 congruent to 2 mod 13?? – Ian Murphy Oct 28 '14 at 19:23
We have $3^3\equiv 1$, so since $3$ divides $2019$, the last part is congruent to $1$, giving sum congruent to $0$. – André Nicolas Oct 28 '14 at 19:23
Sorry, a silly mistake. Fixed. – ajotatxe Oct 28 '14 at 23:32

Modular Arithmetic Involving Remainders

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