Find the principal remainder of $\frac{431^5 + 611}{27}$

Question

I have this problem: $\frac{431^5 + 611}{27}$

I'm supposed to find the principal remainder by hand. But I have no idea how to start when $431$ has en exponent of $5$. Can someone please explain? Thanks.

Answer 1

We write the fraction as $\dfrac{(15\times27+27-1)^{5}+21\times27+44}{27}$

which is equal to $\dfrac{M\times\,27\,-1+44}{27}=M+\dfrac{43}{27}$

and $43=27+16$, hence the remainder is $16$.

I am sorry I have not checked the result with a calculator.