Question

O restante quando o determinante da matriz

$\begin{pmatrix}\ \ {2014}^{2014} & {2015}^{2015} & {2016}^{2016} \\ {2017}^{2017} & {2018}^{2018} & {2019}^{2019} \\ {2020}^{2020} & {2021}^{2021} & {2022}^{2022} \ \end{pmatrix}$

é dividido por 5 é:

( Gabarito: 4 )

#Cálculo e explicação

Answer 1

$\left(\begin{array}{ccc}2014^{2014}&2015^{2015}&2016^{2016}\\2017^{2017}&2018^{2018}&2019^{2019}\\2020^{2020}&2021^{2021}&2022^{2022}\end{array}\right)$

Pela Regra de Sarrus:

$\text{Det}=\left|\begin{array}{ccc}2014^{2014}&2015^{2015}&2016^{2016}\\2017^{2017}&2018^{2018}&2019^{2019}\\2020^{2020}&2021^{2021}&2022^{2022}\end{array}\right|\begin{array}{cc}2014^{2014}&2015^{2015}\\2017^{2017}&2018^{2018}\\2020^{2020}&2021^{2021}\end{array}\right|$

$\bullet~\text{Diagonais}~\text{principais}$

$2014^{2014}\cdot2018^{2018}\cdot2022^{2022}+2015^{2015}\cdot2019^{2019}\cdot2020^{2020}+\\2016^{2016}\cdot2017^{2017}\cdot2021^{2021}$

As congruências de mesmo módulo somam-se e subtraem-se membro a membro tal qual as igualdades. Desse modo, vamos determinar o resto de cada uma dessas parcelas na divisão por $5$ e depois somá-los.

Pelo Pequeno Teorema de Fermat, $a^{p-1}\equiv1\pmod{p}$ se $p$ é primo e $a$ é um inteiro não divisível por $p$

$\star~~2014^{2014}\cdot2018^{2018}\cdot2022^{2022}$

Note que $2014\equiv4\pmod{5}$. Assim, $2014^{2014}\equiv4^{2014}\pmod{5}$

Pelo Pequeno Teorema de Fermat, $4^{4}\equiv1\pmod{5}$. Como $2014=4\cdot503+2$, temos que:

$4^{2014}\equiv(4^{4})^{503}\cdot4^{2}\equiv1^{503}\cdot16\equiv1\cdot16\equiv1\pmod{5}$. Logo, $2014^{2014}\equiv1\pmod{5}$

Analogamente, temos que $2018\equiv3\pmod{5}$. Desse modo, $2018^{2018}\equiv3^{2018}\pmod{5}$

O Pequeno Teorema de Fermat nos garante que $3^{4}\equiv1\pmod{5}$. Note que $2018=4\cdot504+2$

Assim, $3^{2018}\equiv(3^{4})^{504}\cdot3^{2}\equiv1^{504}\cdot9\equiv1\cdot9\equiv4\pmod{5}$

Portanto, $2018^{2018}\equiv4\pmod{5}$

Temos que $2022\equiv2\pmod{5}$, então $2022^{2022}\equiv2^{2022}\pmod{5}$

Pelo Pequeno Teorema de Fermat, $2^{4}\equiv1\pmod{5}$. Veja que $2022=4\cdot505+2$

Desta maneira, $2^{2022}\equiv(2^{4})^{505}\cdot2^{2}\equiv1^{505}\cdot4\equiv1\cdot4\equiv4\pmod{5}$

Assim, $2022^{2022}\equiv4\pmod{5}$. 

Logo, $2014^{2014}\cdot2018^{2018}\cdot2022^{2022}\equiv1\cdot4\cdot4\equiv16\equiv1\pmod{5}$

$\star~~2015^{2015}\cdot2019^{2019}\cdot2020^{2020}$

Como $2015\equiv0\pmod{5}$, temos que $2015^{2015}\equiv0^{2015}\equiv0\pmod{5}$

Analogamente, $2020\equiv0\pmod{5}$ e, por consequência, $2020^{2020}\equiv0^{2020}\equiv0\pmod{5}$

Observe que $2019\equiv4\pmod{5}$, de modo que $2019^{2019}\equiv4^{2019}\pmod{5}$

Pelo Pequeno Teorema de Fermat, $4^{4}\equiv1\pmod{5}$. Como $2019=4\cdot504+3$, temos que:

$4^{2019}\equiv(4^{4})^{504}\cdot4^{3}\equiv1^{504}\cdot64\equiv1\cdot64\equiv4\pmod{5}$

Assim, $2019^{2019}\equiv4\pmod{5}$

Logo, $2015^{2015}\cdot2019^{2019}\cdot2020^{2020}\equiv0\cdot4\cdot0\equiv0\pmod{5}$

$\star~~2016^{2016}\cdot2017^{2017}\cdot2021^{2021}$

Como $2016\equiv1\pmod{5}$, segue que $2016^{2016}\equiv1^{2016}\equiv1\pmod{5}$

Analogamente, $2021\equiv1\pmod{5}$. Então, $2021^{2021}\equiv1^{2021}\equiv1\pmod{5}$

Note que $2017\equiv2\pmod{5}$. Desse modo, $2017^{2017}\equiv2^{2017}\pmod{5}$

O Pequeno Teorema de Fermat nos garante que $2^{4}\equiv1\pmod{5}$. Veja que $2017=4\cdot504+1$

Desta maneira, $2^{2017}\equiv(2^{4})^{504}\cdot2^{1}\equiv1^{504}\cdot2\equiv1\cdot2\equiv2\pmod{5}$

Logo, $2017^{2017}\equiv2\pmod{5}$

Portanto, $2016^{2016}\cdot2017^{2017}\cdot2021^{2021}\equiv1\cdot2\cdot1\equiv2\pmod{5}$

$\bullet~\text{Diagonais}~\text{secund}\acute{\text{a}}\text{rias}$

$2016^{2016}\cdot2018^{2018}\cdot2020^{2020}+2014^{2014}\cdot2019^{2019}\cdot2021^{2021}+\\2015^{2015}\cdot2017^{2017}\cdot2022^{2022}$

Sabemos que:

$2014^{2014}\equiv1\pmod{5}~~~~~~~~~~~~~~~2019^{2019}\equiv4\pmod{5}~\\2015^{2015}\equiv0\pmod{5}~~~~~~~~~~~~~~~2020^{2020}\equiv0\pmod{5}~\\2016^{2016}\equiv1\pmod{5}~~~~~~~~~~~~~~~2021^{2021}\equiv1\pmod{5}~\\2017^{2017}\equiv2\pmod{5}~~~~~~~~~~~~~~~2022^{2022}\equiv4\pmod{5}~\\2018^{2018}\equiv4\pmod{5}$

Assim, podemos afirmar que:

$\star~~2016^{2016}\cdot2018^{2018}\cdot2020^{2020}\equiv1\cdot4\cdot0\equiv0\pmod{5}$

$\star~~2014^{2014}\cdot2019^{2019}\cdot2021^{2021}\equiv1\cdot4\cdot1\equiv4\pmod{5}$

$\star~~2015^{2015}\cdot2017^{2017}\cdot2022^{2022}\equiv0\cdot2\cdot4\equiv0\pmod{5}$

Logo:

$\text{Det}\equiv1+0+2-0-4-0\equiv-1\equiv5-1\equiv4\pmod{5}$

Portanto, o resto da divisão é $4$