Question

What is the remainder when 81x to the power of 1857 is divided with x+1

Answer 1

We have to find remainder when $(81 x)^{1857}$  divided by x+1 .

$(81 x)^{1857}= (81)^{1857} \times x^{1857}$

⇒$(81)^{1857}$ is a constant term.By taking it out,then dividing $x^{1857}$ by x+1, we get

Put x+1 =0

x=-1

Let p(x)=$(81 x)^{1857}$ = $(81)^{1857} \times x^{1857}$

P(-1)=$(81)^{1857} \times (-1)^{1857}=(81)^{1857} \times (-1)= -(81)^{1857}$   ⇒[ (-1) raised odd integer=-1]

$-(81)^{1857}$ is the remainder when $(81 x)^{1857}$  divided by x+1 .