Question 727402
Dividing 287 by 3 we get 95 for a quotient and {{{highlight(remainder=red(2))}}}.
Dividing 95 by 3 we get 31 for a quotient and {{{highlight(remainder=green(2))}}}.
Dividing 31 by 3 we get 10 for a quotient and {{{highlight(remainder=red(1))}}}.
Dividing 10 by 3 we get 3 for a quotient and {{{highlight(remainder=blue(1))}}}.
Dividing 3 by 3 we get {{{highlight(green(1))}}} for a quotient and {{{highlight(remainder=0)}}}.
 
The answer is {{{highlight(101122)=green(1)}}}{{{0}}}{{{blue(1)}}}{{{red(1)}}}{{{green(2)}}}{{{red(2))}}} in base 3, whose value is
{{{green(1)*3^5+0*3^4+blue(1)*3^3+red(1)*3^2+green(2)*3+red(2)=287}}}
 
The reason why the repeated division works can be explained through that example like this
The last division tells us that
{{{red(3)=3*green(1)+0}}} because 3 ÷ 3 = 1 R0
The previous division tells us that
{{{10=3*red(3)+blue(1)}}} because 10 ÷ 3 = 3 R1
Putting them both together
{{{10=3*(3*green(1)+0)+blue(1)=green(1)*3^2+0*3+blue(1)}}}
The division before that told us that
{{{31=3*10+red(1)}}} and that along with
{{{10=green(1)*3^2+0*3+blue(1)}}} gets us
{{{31=3(green(1)*3^2+0*3+blue(1))+red(1)}}}={{{green(1)*3^3+0*3^2+blue(1)*3+red(1)}}}
The second division told us that
{{{95=3*31+green(2)}}} and that along with
{{{31=green(1)*3^3+0*3^2+blue(1)*3+red(1)}}} gets us
{{{95=3*(green(1)*3^3+0*3^2+blue(1)*3+red(1))+green(2)}}}=
{{{green(1)*3^4+0*3^3+blue(1)*3^2+red(1)*3+green(2)}}}
The first division told us that
{{{287=3*95+red(2)}}} and that along with
{{{95=green(1)*3^4+0*3^3+blue(1)*3^2+red(1)*3+green(2)}}} gets us
{{{287=3*(green(1)*3^4+0*3^3+blue(1)*3^2+red(1)*3+green(2))+red(2)}}} which means
{{{287=green(1)*3^5+0*3^4+blue(1)*3^3+red(1)*3^2+green(2)*3+red(2)}}}