



 
Hi, I am going to make liberal us of the Binomial Theorem which says
What is important in what I am going to do is that \[\left(\begin{array}{c}n\\k\end{array}\right) \mbox{ is a positive integer for } k = 0, \cdot \cdot \cdot ,n\] and that \[\left( \begin{array}{c}n\\n\end{array}\right) = 1\] For $2018^{2019}$ I want to write $2018$ as $2016 + 2$ and think about expanding \[2018^{2019} = (2016 + 2)^{2019}\] using the Binomial Theorem with $a = 2016$ and $b = 2.$ Since $2016$ is divisible by $4,$ each of the terms on the right side of the Binomial Theorem is divisible by $4$ except perhaps the last term. Hence if $2018^{2019}$ is divided by $4$ the remainder is \[\left(\begin{array}{c}n\\n \end{array}\right) b^{n} = 1 \times 2^{2019}.\] But \[2^{2019} = (2 \times 2)^{1009} \times 2 = 4^{1009} \times 2\] and division of this term by $4$ leaves a remainder of $2.$ Hence when $2018^{2019}$ is divided by $4$ the remainder is $2.$ In a similar fashion for $2019^{2018}$ write $2019$ as $2016 + 3$ and again use the Binomial Theorem. As before dividing by $4$ leaves a remainder of $3^{2018}.$ This time write $3 = 4  1$ and apply the Binomial Theorem to $(4  1)^{2018}.$ All but the last term is divisible by $4$ and the last term is $(1)^{2018} = 1.$ hence when $2019^{2018}$ is divided by $4$ the reminder is $1.$ Finally we see that when $2018^{2019}  2019^{2018}$ is divided by $4$ the remainder is $2  1 = 1.$ Harley 



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